MATHS MCQ

But $x$ is real, therefore ${B^2} - 4AC \ge 0$

$ \Rightarrow {\cos ^2}\theta \ge 4(1)(1) \Rightarrow {\cos ^2}\theta \ge 4$, which is impossible.

$ \Rightarrow {\sin ^2}\theta = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $

$\Rightarrow {(a + b)^2} - 4ab \le 0$

$ \Rightarrow {(a - b)^2} \le 0 $

$\Rightarrow a = b.$

${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})$

$ \Rightarrow \,3\,(1 - {m^2})\,{t^2} + (4{m^2} + 6m - 4)\,t - 8m = 0$

$ \Rightarrow \,(3t - 4)\,[(1 - {m^2})\,t + 2m] = 0$,

which is true, if $t = \tan \theta = \frac{4}{3}$ or $\tan \theta = \frac{{2m}}{{{m^2} - 1}}$.

${\cos ^2}A = \sin A\,\tan A = \frac{{{{\sin }^2}A}}{{\cos A}}\, $

$\Rightarrow \,{\cos ^3}A - {\sin ^2}A = 0$

Hence ${\cos ^3}A + {\cos ^2}A = {\sin ^2}A + {\cos ^2}A = 1$

$\therefore \,\,\,\sec \theta - \tan \theta = {e^{ - x}}$…..$(ii)$

From $(i)$ and $(ii),$

$\,2\sec \theta = {e^x} + {e^{ - x}}\,$

$\Rightarrow \,\cos \theta = \frac{2}{{{e^x} + {e^{ - x}}}}.$

$ \Rightarrow \,\cos \theta = (\sqrt 2 + 1)\,\sin \theta \, $

$\Rightarrow \,(\sqrt 2 - 1)\cos \theta = \sin \theta $

$ \Rightarrow \,\sqrt 2 \,\cos \theta - \cos \theta = \sin \theta $

$\Rightarrow \,\sin \theta + \cos \theta = \sqrt 2 \,\cos \theta .$

$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}$

$ = \frac{{2\,\,\sin \frac{A}{2}\,\left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)}}{{2\,\,\cos \frac{A}{2}\,\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}}$

$= \tan \frac{A}{2}$. 

Trick : Put $A = {60^o}.$ 

$\frac{{1 + (\sqrt 3 /2) - (1/2)}}{{1 + (\sqrt 3 /2) + (1/2)}} = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} = \frac{1}{{\sqrt 3 }}$ 

which is given by option $(c)$,

$i.e.$ $\tan \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$

$\frac{{1 + \cos y - {{\sin }^2}y}}{{1 + \cos y}} + \frac{{(1 - {{\cos }^2}y) - {{\sin }^2}y}}{{\sin y\,(1 - \cos y)}}$

$ = \frac{{\cos y\,(1 + \cos y)}}{{1 + \cos y}} + 0 = \cos y.$

and $2x\,\sec \theta - y\,\,{\rm{cosec}}\,\theta = 3$…..$(ii)$

$ \Rightarrow \,\,\frac{{2x}}{{\cos \theta }} - \frac{y}{{\sin \theta }} = 3$

$ \Rightarrow \,\,2x\,\sin \theta - y\,\cos \theta - 3\,\sin \theta \cos \theta = 0$…..$(iii)$

Solving $(i)$ and $(iii)$, 

we get $y = \sin \theta $ and $x = 2\,\,\cos \theta $

Now, ${x^2} + 4{y^2} = 4\,\,{\cos ^2}\theta + 4\,\,{\sin ^2}\theta $ 

$ = 4\,({\cos ^2}\theta + {\sin ^2}\theta ) = 4$.

$ = \frac{{1 - \sin \,\phi}}{{\cos \,\phi}}\,.\,\frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$ 

$ \Rightarrow \,xy + 1 = \frac{{1 - \sin \,\phi+ \cos \,\phi- \sin \,\phi\,\cos \,\phi+ \sin \phi \cos \phi}}{{\cos \phi\sin \phi}}$

$= \frac{{1 - \sin \,\phi+ \cos \,\phi}}{{\cos \,\phi\sin \,\phi}}$…..$(i)$ 

$x - y = (\sec \,\phi- \tan \,\phi) - (\cos ec\,\phi+ \cot \,\phi)$ $ = \frac{{1 - \sin \,\phi}}{{\cos \,\phi}} - \frac{{1 + \cos \,\phi}}{{\sin \,\phi}}$

$= \frac{{\sin \,\phi- {{\sin }^2}\phi- \cos \,\phi- {{\cos }^2}\phi}}{{\cos \,\phi\,\sin \,\phi}}$

$ = \frac{{\sin \,\phi - \cos \,\phi- 1}}{{\cos \,\phi \,\sin \,\phi}}$…..$(ii)$

Adding $(i)$ and $(ii)$ we get, $xy + 1 + (x - y) = 0$ 

$ \Rightarrow x = \frac{{y - 1}}{{y + 1}}$.

$ \Rightarrow \,\,p + q = \frac{{\cos \theta }}{{1 + \sin \theta }} + \frac{{2\,\sin \theta }}{{1 + \sin \theta  + \cos \theta }}\,$

$\Rightarrow \,p + q = 1.$

and $\cos \,2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \frac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}$

$\therefore $ $\sin \theta = \pm \frac{a}{{\sqrt {{a^2} + {b^2}} }},\,\,\cos \,\theta = \pm \frac{b}{{\sqrt {{a^2} + {b^2}} }}$

Now, $\frac{{\sin \,\theta }}{{\cos {\,^8}\theta }} + \frac{{\cos \,\theta }}{{{{\sin }^8}\theta }}$

$= \frac{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}} + \frac{{\left( {\frac{b}{{\sqrt {{a^2} + {b^2}} }}} \right)}}{{{{\left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right)}^8}}}$

$ = \frac{{a\,{{({a^2} + {b^2})}^4}}}{{{b^8}\,{{({a^2} + {b^2})}^{1/2}}}} + \frac{{b\,{{({a^2} + {b^2})}^4}}}{{{a^8}\,{{({a^2} + {b^2})}^{1/2}}}}$

$ = \pm \frac{{{{({a^2} + {b^2})}^4}}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\frac{a}{{{b^8}}} + \frac{b}{{{a^8}}}} \right)$.

$\Rightarrow \,\,\cos x = {\sin ^2}x$

$\therefore \,\,{\sin ^2}x + {\sin ^4}x = \cos x + {\cos ^2}x = 1$.

$\Rightarrow \,\,\sin x = {\cos ^2}x$

$\therefore$ ${\cos ^8}x + 2{\cos ^6}x + {\cos ^4}x $

$= {\sin ^4}x + 2{\sin ^3}x + {\sin ^2}x$

$ = {(\sin x + {\sin ^2}x)^2} = 1$.

and $x\,\sin \,\alpha - y\,\cos \,\alpha = 0$…..$(ii)$

Now from $(ii)$, $x\,\sin \,\alpha = y\,\cos \,\alpha $

Putting in $(i),$ we get

$ \Rightarrow \,\,y\,\cos \alpha \,{\sin ^2}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \alpha \,\left\{ {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right\} = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \,\alpha = \sin \,\alpha \,\cos \,\alpha \, $

$\Rightarrow \,\,y = \sin \,\alpha $ and $x = \cos \,\alpha $

Hence, ${x^2} + {y^2} = {\sin ^2}\alpha + {\cos ^2}\alpha = 1.$

and $\sin \theta + \cos \theta = b$…..$(ii)$

Now ${({b^2} - 1)^2}({a^2} + 4)$

$ = {\left\{ {{{(\sin \theta + \cos \theta )}^2} - 1} \right\}^2}\left\{ {{{(\tan \theta - \cot \theta )}^2} + 4} \right\}$

$ = {[1 + \sin 2\theta - 1]^2}[{\tan ^2}\theta + {\cot ^2}\theta - 2 + 4]$

$ = {\sin ^2}2\theta ({\rm{cose}}{{\rm{c}}^2}\theta + {\sec ^2}\theta )$

$ = 4{\sin ^2}\theta {\cos ^2}\theta \left[ {\frac{1}{{{{\sin }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right] = 4$. 

Trick : Obviously the value of expression ${({b^2} - 1)^2}({a^2} + 4)$ is independent of $\theta $,

therefore put any suitable value of $\theta $.

Let $\theta = 45^\circ $, we get $a = 0,\;b = \sqrt 2 $

so that ${[{(\sqrt 2 )^2} - 1]^2}$ $({0^2} + 4) = 4.$

and $\cos 22\frac{{{1^o}}}{2} = \frac{1}{2}\sqrt {2 + \sqrt 2 } $

$\therefore \left( {1 + \cos 22\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 67\frac{{{1^o}}}{2}} \right)\,\left( {1 + \cos 112\frac{{{1^o}}}{2}} \right)$

$\left( {1 + \cos 157\frac{{{1^o}}}{2}} \right)$

$ = \left( {1 + \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)\,\left( {1 + \frac{1}{2}\sqrt {2 - \sqrt 2 } } \right)\,\left( {1 - \frac{1}{2}\sqrt {2 - \sqrt 2 } } \right)\,$

$\left( {1 - \frac{1}{2}\sqrt {2 + \sqrt 2 } } \right)$

$ = \left[ {1 - \frac{1}{4}(2 + \sqrt 2 )} \right]\,\left[ {1 - \frac{1}{4}(2 - \sqrt 2 )} \right]$

$ = \frac{{(4 - 2 - \sqrt 2 )(4 - 2 + \sqrt 2 )}}{{16}}$

$ = \frac{{(2 - \sqrt 2 )(2 + \sqrt 2 )}}{{16}} = \frac{{4 - 2}}{{16}} = \frac{1}{8}$.

$ = 6[{({\sin ^2}\theta + {\cos ^2}\theta )^3} - 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )]$

$ - 9[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}] + 4$

$ = 6[1 - 3{\sin ^2}\theta {\cos ^2}\theta ] - 9\,[1 - 2{\sin ^2}\theta {\cos ^2}\theta ] + 4$

$ = 6 - 9 + 4 = 1$.

$\sin \theta = \frac{1}{{\sqrt {50} }},\,\,\cos \theta = \frac{7}{{\sqrt {50} }},\,\,\cos \phi = \frac{3}{{\sqrt {10} }}$

$\therefore \,\,\cos 2\phi  = 2{\cos ^2}\phi - 1 = 2.\frac{9}{{10}} - 1 = \frac{8}{{10}}$

$\sin 2\phi = 2\sin \phi \cos \phi = 2 \times .\frac{1}{{\sqrt {10} }} \times \frac{3}{{\sqrt {10} }} = \frac{6}{{10}}$

$\therefore \cos (\theta + 2\phi ) = \cos \theta \cos 2\phi - \sin \theta \sin 2\phi $

$ = \frac{7}{{\sqrt {50} }} \times \frac{8}{{10}} - \frac{1}{{\sqrt {50} }}.\frac{6}{{10}}$

$ = \frac{{56 - 6}}{{10\sqrt {50} }} = \frac{{50}}{{10\sqrt {50} }} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}$

$\therefore \theta + 2\phi = {45^o}$.

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ..... + {\sin ^2}{85^o} + {\sin ^2}{90^o}.$

We know that $\sin {90^o} = 1$ or ${\sin ^2}{90^o} = 1$. 

Similarly, $\sin {45^o} = \frac{1}{{\sqrt 2 }}{\rm{or}}\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{45^o} = \frac{1}{2}$ 

and the angles are in $A.P. $ of $18$ terms. 

We also know that ${\sin ^2}{85^o} = {[\sin ({90^o} - {5^o})]^2}$$ = {\cos ^2}{5^o}.$ 

Therefore from the complementary rule, we find 

$\therefore$ ${\sin ^2}{5^o} + {\sin ^2}{85^o} = {\sin ^2}{5^o} + {\cos ^2}{5^o} = 1.$ 

Therefore, 

${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + {\sin ^2}{85^o} + {\sin ^2}{90^o}$ 

$ = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 + \frac{1}{2} = 9\frac{1}{2}$.

$\therefore \,\,\,\tan \left( {\frac{{A + B}}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right)$ 

==> $\frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}.\tan \frac{B}{2}}} = \cot \frac{C}{2} $

$\Rightarrow \frac{{\frac{1}{3} + \frac{2}{3}}}{{1 - \frac{1}{3}.\frac{2}{3}}} = \frac{9}{7} = \cot \frac{C}{2}$ 

$\therefore  \,\,  \tan \frac{C}{2} = \frac{7}{9}$.

==> ${\cos ^2}x + {\sin ^2}x + 2\cos x\sin x + k\sin x\cos x - 1 = 0,\,\forall \,x$

==> $(k + 2)\cos x\sin x = 0,\,\forall \,x$ ==> $k + 2 = 0$ ==> $k = - 2$.

and $\sec (A + B) = \frac{2}{{\sqrt 3 }} \Rightarrow A + B = \frac{{11\pi }}{6}$…..(ii)

from (i) and (ii),

$ \Rightarrow B = \frac{{19\pi }}{{24}}$.

$\therefore \quad \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=1$

$\Rightarrow \quad \theta_{1}=\theta_{2}=\theta_{3}=\frac{\pi}{2}$

$ \therefore \quad \cos \theta_{1}=\cos \theta_{2}=\cos \theta_{3}=0 $

$ \therefore \quad \cos \theta_{1} =\cos \theta_{2}=\cos \theta_{3}=0 $

$=\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\tan 40^{\circ} \tan 20^{\circ}} $

$ \therefore  \sqrt{3}-\sqrt{3} \tan 40^{\circ} \tan 20^{\circ}=\tan 40^{\circ}+\tan 20^{\circ} $

Hence $\tan {40^\circ } + \tan {20^\circ } + \sqrt 3 \tan {40^\circ }\tan {20^\circ }$

$=\sqrt{3}$

$ \Rightarrow \,\sin x + 3\cos x = 3\cos y - \sin y$…..$(i)$

$ \Rightarrow \,r\cos \,(x - \alpha ) = r\cos \,(y + \alpha ),$

where $r = \sqrt {10} ,\,\tan \alpha = \frac{1}{3}$

$ \Rightarrow \,x - \alpha = \pm (y + \alpha )\, $

$\Rightarrow \,x = - y$ or $x + y = 2\alpha $

Clearly, $x = - y$ satisfies $(i); $

$\therefore \;\frac{{\sin \,3x}}{{\sin \,3y}} = \frac{{ - \sin \,3y}}{{\sin \,3y}} = - 1$.

which is zero, if, $\sin \left( {\frac{\pi }{4} + x - y} \right) = 0$

$i.e.$, $\frac{\pi }{4} + x - y = n\pi (n \in I)$

$\Rightarrow x = n\pi - \frac{\pi }{4} + y$.

$= \frac{{1 - \cos \alpha + 1 + \cos \alpha }}{{\sqrt {1 - {{\cos }^2}\alpha } }}$

$ = \frac{2}{{ \pm \sin \alpha }}$

$ = \frac{2}{{ - \sin \alpha }},\,\,\left( {{\rm{since \,\,}}\pi < \alpha < \frac{{{\rm{3}}\pi }}{{\rm{2}}}} \right).$

$ \Rightarrow \,\,\frac{{2\,\,\sin \frac{{A + B}}{2}.\cos \frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2}.\cos \frac{{A - B}}{2}}} = \frac{C}{D}$

$ \Rightarrow \,\,\tan \frac{{A + B}}{2} = \frac{C}{D}$

Thus, $\sin \,(A + B) = \frac{{2\,\,\tan \frac{{A + B}}{2}}}{{1 + {{\tan }^2}\frac{{A + B}}{2}}}$

$ = \frac{{2\,\frac{C}{D}}}{{1 + \frac{{{C^2}}}{{{D^2}}}}} = \frac{{2CD}}{{({C^2} + {D^2})}}$.

$\frac{{\sin A}}{{\sin B}} = \frac{{\cos A}}{{\cos B}}\,$

$ \Rightarrow \,\,\sin A\,\cos B - \cos A\,\sin B = 0$

$ \Rightarrow \,\,\sin \,(A - B) = 0$

Hence, $\sin \,\left( {\frac{{A - B}}{2}} \right) = 0.$

$\tan \,({20^o} + {40^o}) = \frac{{\tan \,{{20}^o} + \tan \,{{40}^o}}}{{1 - \tan \,{{20}^o}\,\tan \,{{40}^o}}}$

$ \Rightarrow \,\,\,\sqrt 3 = \frac{{\tan \,{{20}^o} + \tan \,{{40}^o}}}{{1 - \tan \,{{20}^o}\,\tan \,{{40}^o}}}$

$ \Rightarrow \,\,\sqrt 3 - \sqrt 3 \,\tan \,{20^o}\,\tan \,{40^o} = \tan \,{20^o} + \tan \,{40^o}$

$ \Rightarrow \,\,\tan \,{20^o} + \tan \,{40^o} + \sqrt 3 \,\tan \,{20^o}\,\tan \,{40^o} $

$= \sqrt 3 .$

$= \frac{1}{{(1 + \tan A)\,(1 + \tan B)}}$ $ = \frac{1}{{\tan A + \tan B + 1 + \tan A\tan B}}$                                                                                  $[\,\because \tan (A + B) = \tan {225^o}]$

$ \Rightarrow \,\tan \,A + \tan \,B = 1 - \tan \,A\,\tan B$

$ = \frac{1}{{1 - \tan A\,\tan B + 1 + \tan A\tan B}} $

$= \frac{1}{2}$.

Now, $\cos \,(A + B) = \cos A\,\cos B - \sin A\,\sin B$

$ = \sqrt {1 - \frac{{16}}{{25}}} \,\left( { - \frac{{12}}{{13}}} \right) - \frac{4}{5}\sqrt {1 - \frac{{144}}{{169}}} $

$ = - \frac{3}{5} \times \frac{{12}}{{13}} - \frac{4}{5}\,\left( { - \frac{5}{{13}}} \right)$

$= - \frac{{16}}{{65}}$

(Since $A$ lies in first quadrant અને $B$ lies in third quadrant).

$\Rightarrow \,\tan \,(A + B) = \tan \,\frac{\pi }{4}$

$ \Rightarrow \,\,\frac{{\tan A + \tan B}}{{1 - \tan A\,\tan B}} = 1$ 

$ \Rightarrow \,\,\tan A + \tan B + \tan A\,\tan B = 1$

$ \Rightarrow \,\,(1 + \tan A)\,(1 + \tan B) = 2$.

$ \Rightarrow \,\,\sin \theta = \sqrt {1 - \frac{{{8^2}}}{{{{17}^2}}}} = \frac{{15}}{{17}}$ 

The value of the given expression

$ = \cos \,\,{30^o}\,.\,\cos \theta - \sin \,\,{30^o}\sin \theta + \cos \,\,{45^o}\cos \theta $ 

$ + \sin \,\,{45^o}\sin \theta + \cos \,\,{120^o}\cos \theta + \sin \,\,{120^o}\sin \theta $

$ = \cos \theta \,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right) - \sin \theta \,\left( {\frac{1}{2} - \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2}} \right)$ 

$ = \frac{8}{{17}}\,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right) + \frac{{15}}{{17}}\,\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} - \frac{1}{2}} \right)$

$ = \frac{{23}}{{17}}\,\left( {\frac{{\sqrt 3 - 1}}{2} + \frac{1}{{\sqrt 2 }}} \right)$.

$= 2 sin 54^\circ cos 7^\circ - 2 sin 18^\circ cos 7^\circ$

$ = \,\,2\,\,\cos \,\,{7^o}\,(\sin \,\,{54^o} - \sin \,\,{18^o})$

$ = \,\,2\,\,\cos \,\,{7^o}\,\,.\,\,2\,\,\cos \,\,{36^o}\,\,.\,\,\sin \,\,{18^o}$

$ = \,\,4.\,\cos \,\,{7^o}.\,\frac{{\sqrt 5 + 1}}{4}.\frac{{\sqrt 5 - 1}}{4} = \cos \,\,{7^{o.}}$.

$ \Rightarrow \cos \alpha = \frac{8}{{17}},\sin \beta = \frac{{12}}{{13}}$

and $\cos \beta = - \frac{5}{{13}}$

==> $\pi < \beta < \frac{{3\pi }}{2}$, 

$\therefore \cos \beta = - \frac{5}{{13}}$

$\sin (\beta - \alpha ) = \sin \beta \cos \alpha - \cos \beta \sin \alpha $ = $\frac{{171}}{{221}}$.

==> $\tan ({70^o} - {20^o}) = \frac{{\tan {{70}^o} - \tan {{20}^o}}}{{1 + \tan {{70}^o}\tan {{20}^o}}}$

==> $\tan {50^o} + \tan {70^o}\tan {20^o}\tan {50^o} = \tan {70^o} - \tan {20^o}$

==> $\tan {50^o} + \tan {50^o} = \tan {70^o} - \tan {20^o}$

                                               $[\,\because \tan {70^o} = \cot {20^o}]$

==> $2\tan {50^o} + \tan {20^o} = \tan {70^o}$

==> $2\tan {50^o} + \tan {20^o} = \tan {50^o} + \sec {50^o}$.

$ = 2\cos \frac{\pi }{{13}}.\cos \frac{{9\pi }}{{13}} + 2\cos \frac{{4\pi }}{{13}}\cos \frac{\pi }{{13}}$

$ = 2\cos \frac{\pi }{{13}}\left[ {\cos \frac{{9\pi }}{{13}} + \cos \frac{{4\pi }}{{13}}} \right]$ 

$ = 2\cos \frac{\pi }{{13}}\left[ {2\cos \frac{\pi }{2}.\cos \frac{{5\pi }}{{26}}} \right] = 0$,.                  $\left[ \because {\cos \frac{\pi }{2} = 0} \right]$

$ \Rightarrow \,\,\frac{{m + 1}}{{m - 1}} = \frac{{\cos A + \cos B}}{{\cos A - \cos B}} $

$= \frac{{2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{B - A}}{2}} \right)}}{{2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{B - A}}{2}} \right)}}$

$ = \cot \,\left( {\frac{{A + B}}{2}} \right)\,\cot \,\left( {\frac{{B - A}}{2}} \right)$

Hence, $\cot \,\left( {\frac{{A + B}}{2}} \right) = \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$.

$ = \frac{1}{4}\,(2\,\,\sin \,\,{12^o}\,\sin \,\,{48^o})\,\,(2\,\,\sin \,\,{24^o}\,\,\sin \,\,{84^o})$

$ = \frac{1}{2}(\cos \,\,{36^o} - \cos \,\,{60^o})\,\,(\cos \,\,{60^o} - \cos \,\,{108^o})$

$ = \frac{1}{4}\,\left( {\cos \,\,{{36}^o} - \frac{1}{2}} \right)\,\,\left( {\frac{1}{2} + \sin \,\,{{18}^o}} \right)$

$ = \frac{1}{4}\left\{ {\frac{1}{4}(\sqrt 5 + 1) - \frac{1}{2}} \right\}\,\left\{ {\frac{1}{2} + \frac{1}{4}(\sqrt 5 - 1)} \right\} = \frac{1}{{16}}$

and $\cos \,\,{20^o}\,\cos \,\,{40^o}\,\,\cos \,\,60\,\,\cos \,\,{80^o}$

$ = \frac{1}{2}[\cos \,({60^o} - {20^o})\,\cos \,\,{20^o}\,\cos \,({60^o} + {20^o})]$

$ = \frac{1}{2}\,\left[ {\frac{1}{4}\cos \,\,3\,\,({{20}^o})} \right] = \frac{1}{8}\cos \,\,{60^o} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{{16}}$.

$ = 1 - {\sin ^2}\left( {\frac{\pi }{{12}}} \right) + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\left( {\frac{{5\pi }}{{12}}} \right)$

$ = 1 + \frac{1}{2} + \left( {{{\cos }^2}\frac{{5\pi }}{{12}} - {{\sin }^2}\frac{\pi }{{12}}} \right)$

$ = \frac{3}{2} + \cos \left( {\frac{{5\pi }}{{12}} + \frac{\pi }{{12}}} \right)\cos \left( {\frac{{5\pi }}{{12}} - \frac{\pi }{{12}}} \right) $

$= \frac{3}{2} + \cos \frac{\pi }{2}\cos \frac{\pi }{3}$

$ = \frac{3}{2} + 0.\frac{1}{2} = \frac{3}{2}$.

$ = \frac{1}{4}\left[ {2\sin \frac{\pi }{{16}}\sin \frac{{3\pi }}{{16}}.2\sin \frac{{5\pi }}{{16}}\sin \frac{{7\pi }}{{16}}} \right]$

$ = \frac{1}{4}\left[ {\left( {\cos \frac{\pi }{8} - \cos \frac{\pi }{4}} \right)\left( {\cos \frac{\pi }{8} - \cos \frac{{3\pi }}{4}} \right)} \right]$

$ = \frac{1}{4}\left[ {\left( {\cos \frac{\pi }{8} - \frac{1}{{\sqrt 2 }}} \right)\left( {\cos \frac{\pi }{8} + \frac{1}{{\sqrt 2 }}} \right)} \right]$

$ = \frac{1}{4}\left[ {\left( {{{\cos }^2}\frac{\pi }{8} - \frac{1}{2}} \right)} \right] $

$= \frac{1}{8}\left[ {2{{\cos }^2}\frac{\pi }{8} - 1} \right]$

$ = \frac{1}{8}\left[ {\cos \frac{\pi }{4}} \right] $

$= \frac{1}{8} \times \frac{1}{{\sqrt 2 }} $

$= \frac{{\sqrt 2 }}{{16}}$.

$ = \frac{1}{2}\left[ {1 + \cos {{152}^o} + 1 + \cos {{32}^o} - \cos {{92}^o} - \cos {{60}^o}} \right]$

$ = \frac{1}{2}\left[ {2 - \frac{1}{2} + \cos {{152}^o} + \cos {{32}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + 2\cos {{92}^o}\cos {{60}^o} - \cos {{92}^o}} \right]$

$ = \frac{1}{2}\left[ {\frac{3}{2} + \cos {{92}^o} - \cos {{92}^o}} \right]$

$ = \frac{3}{4}$.

$ = {\cos ^2}\alpha + {\left\{ {\cos \,(\alpha + {{120}^o}) + \cos \,(\alpha - {{120}^o})} \right\}^2}$$ - 2\,\cos \,(\alpha + {120^o})\,\cos \,(\alpha - {120^o})$

$ = {\cos ^2}\alpha + {\left\{ {\,2\,\cos \,\,\alpha \,\cos \,{{120}^o}} \right\}^2} - 2\,\left\{ {{{\cos }^2}\alpha - {{\sin }^2}\,{{120}^o}} \right\}$

$ = {\cos ^2}\alpha + {\cos ^2}\alpha - 2\,{\cos ^2}\alpha + 2\,{\sin ^2}\,{120^o}$

$ = 2{\sin ^2}{120^o} $

$= 2 \times \frac{3}{4} $

$= \frac{3}{2}$.

$ = \frac{{\sin {{20}^o}}}{{\cos {{20}^o}}} - \frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} + 2\tan {50^o}$

$ = \frac{{\sin {{20}^o}\cos {{70}^o} - \cos {{20}^o}\sin {{70}^o}}}{{\cos {{20}^o}\cos {{70}^o}}}$$ + 2\tan {50^o}$

$ = \frac{{\sin ({{20}^o} - {{70}^o})}}{{\frac{1}{2}[\cos ({{70}^o} + {{20}^o}) + \cos ({{70}^o} - {{20}^o})]}}$$ + 2\tan {50^o}$

$ = \frac{{2\sin ( - {{50}^o})}}{{\cos {{90}^o} + \cos {{50}^o}}} + 2\tan {50^o}$

$ = \frac{{ - 2\sin {{50}^o}}}{{0 + \cos {{50}^o}}} + 2\tan {50^o}$

$ = - 2\tan {50^o} + 2\tan {50^o} = 0$.

and $\cos \theta + \cos \phi = b$…..$(ii)$

Squaring, ${\sin ^2}\theta + {\sin ^2}\phi + 2\sin \theta \sin \phi = {a^2}$

and ${\cos ^2}\theta + {\cos ^2}\phi + 2\cos \theta \cos \phi = {b^2}$

Adding, $2+ 2 $$(\sin \theta \sin \phi + \cos \theta \cos \phi ) = {a^2} + {b^2}$

==>$2\cos (\theta - \phi ) = {a^2} + {b^2} - 2$

==> $\cos (\theta - \phi ) = \frac{{{a^2} + {b^2} - 2}}{2}$

$ \Rightarrow \frac{{1 - {{\tan }^2}\frac{{\theta - \phi }}{2}}}{{1 + {{\tan }^2}\frac{{\theta - \varphi }}{2}}} = \frac{{{a^2} + {b^2} - 2}}{2}$

==> $({a^2} + {b^2}) + ({a^2} + {b^2}){\tan ^2}\frac{{\theta - \phi }}{2} - 2 - 2{\tan ^2}\frac{{\theta - \phi }}{2}$

$ = 2 - 2{\tan ^2}\frac{{\theta - \varphi }}{2}$

==>$\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}} = {\tan ^2}\frac{{\theta - \phi }}{2}$

==> $\tan \frac{{(\theta - \phi )}}{2} = \sqrt {\frac{{4 - {a^2} - {b^2}}}{{{a^2} + {b^2}}}} $

Trick : Put $\theta = \frac{\pi }{2},\phi = {0^o}$, then $a = 1 = b$

$\tan \frac{{\theta - \phi }}{2} = 1$, which is given by $(a)$ and $(b).$

Again putting $\theta = \frac{\pi }{4} = \phi $,

we get $\tan \frac{{\theta - \phi }}{2} = 0$, which is given by $(b).$

${=\frac{(1-\sin \alpha)-(1+\sin \alpha)}{\sqrt{1-\sin ^{2} \alpha}}} $

${=\frac{-2 \sin \alpha}{|\cos \alpha|}=\frac{-2 \sin \alpha}{-\cos \alpha}} $

$  \left[\because {\frac{\pi }{2} < \alpha  < \pi \therefore \cos \alpha {\rm{ is  - ve }}} \right] = 2{\rm{tan}}\alpha $

$ = {\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{\pi }{8}$

$ = 2\left( {{{\sin }^2}\frac{\pi }{8} + {{\sin }^2}\frac{{3\pi }}{8}} \right) = 2 \times 1 = 2$.

$ = \tan x + \frac{{\tan x + \sqrt 3 }}{{1 - \sqrt 3 \,\tan x}} + \frac{{\tan x - \sqrt 3 }}{{1 + \sqrt 3 \,\tan x}}$

$ = \tan x + \frac{{8\tan x}}{{1 - 3{{\tan }^2}x}} $

$= \frac{{3\,(3\tan x - {{\tan }^3}x)}}{{1 - 3{{\tan }^2}x}} = 3\tan 3x$

Therefore, the given equation is 

$\Rightarrow$ $3\tan 3x = 3$==> $\tan 3x = 1.$

and $\sin x + \sin y + \sin \alpha = 0.$

The given equation may be written as $\cos x + \cos y = - \cos \alpha $ 

and $\sin x + \sin y = - \sin \alpha .$

Therefore $2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = - \cos \alpha $…..$(i) $

$2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = - \sin \alpha $…..$(ii)$

Divide $(i)$ by $(ii)$, we get 

$\frac{{2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)}}{{2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)}}$$ = \frac{{\cos \alpha }}{{\sin \alpha }}$

==> $\cot \left( {\frac{{x + y}}{2}} \right) = \cot \alpha $.

==> $2\sin 2\theta \cos \theta + \sin 2\theta = \sin \alpha $ 

==> $\sin 2\theta (2\cos \theta + 1) = \sin \alpha $…..$(i)$

Now $\cos \theta + \cos 3\theta + \cos 2\theta = \cos \alpha $

$2\cos 2\,\theta \cos \,\theta + \cos 2\theta = \cos \alpha $

$\cos 2\theta \,(2\cos \theta + 1) = \cos \alpha $…..$(ii) $

From $(i)$ અને $(ii), $

$\tan 2\theta = \tan \alpha $

==> $2\theta = \alpha $ 

==> $\theta = \alpha /2$.

$ = \frac{{\cos {{70}^o} + 2\sin {{140}^o}}}{{\sin {{70}^o}}} $

$= \frac{{\cos {{70}^o} + 2\sin ({{180}^o} - {{40}^o})}}{{\sin {{70}^o}}}$

$ = \frac{{\sin {{20}^o} + \sin {{40}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $

$= \frac{{2\sin {{30}^o}\cos {{10}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}}$

$ = \frac{{\sin {{80}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $

$= \frac{{2\sin {{60}^o}\cos {{20}^o}}}{{\sin {{70}^o}}} = \sqrt 3 $.

$ = \frac{1}{{2\,\,\sin \,\,{{10}^o}}}\,[2\,\,\sin \,\,{10^o}\cos \,\,{10^o}\cos \,\,{20^o}\,\,\cos \,\,{40^o}]$

$ = \frac{1}{{2\,.\,2\,\,\sin \,\,{{10}^o}}}\,[2\,\,\sin \,\,{20^o}\cos \,\,{20^o}\,\,\cos \,\,{40^o}]$

$ = \frac{1}{{2\,.\,4\sin {{10}^o}}}[2\sin {40^o}\cos {40^o}) = \frac{1}{{8\sin {{10}^o}}}(\sin {80^o})$

$ = \frac{1}{{8\,\,\sin \,\,{{10}^o}}}\cos \,\,{10^o} = \frac{1}{8}\cot \,\,{10^o}$.

$ = \tan \,\,{9^o} - \tan \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

$ = (\tan \,\,{9^o} + \cot \,\,{9^o}) - (\tan \,\,{27^o} + \cot \,\,{27^o})$

$ = \frac{{\cos ({9^o} - {9^o})}}{{\sin {9^o}\cos {9^o}}} - \frac{{\cos ({{27}^o} - {{27}^o})}}{{\sin {{27}^o}.\cos {{27}^o}}} = \frac{2}{{\sin {{18}^o}}} - \frac{2}{{\sin {{54}^o}}}$

$ = 2\,\left\{ {\frac{{\sin \,\,{{54}^o} - \sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}\,\sin \,\,{{54}^o}}}} \right\} $

$= 2.\,\,\frac{{2\,.\,\cos \,\,{{36}^o}.\,\sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}.\,\sin \,\,{{54}^o}}} = 4$

$\frac{{(\cos \,6x + \cos \,4x) + 5\,(\cos \,4x + \cos \,2x) + 10\,(\cos \,2x + 1)}}{{\cos \,5x + 5\,\cos \,3x + 10\,\cos x}}$

After solving, we get the required result $i.e.$ $2\,\cos x$.

$=  \frac{{\frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} - \frac{{\sin {{20}^o}}}{{\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{{\frac{{\sin {{70}^o}\cos {{20}^o} - \cos {{70}^o}\sin {{20}^o}}}{{\cos {{70}^o}\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{2}{2} \times \frac{{\sin ({{70}^o} - {{20}^o})\cos {{50}^o}}}{{\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\sin {{50}^o}\cos {{50}^o}}}{{2\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\cos {{50}^o}}}{{\cos {{90}^o} + \cos {{50}^o}}}$

$= \frac{{2\cos {{50}^o}}}{{0 + \cos {{50}^o}}}$

$= 2.$

Therefore $\cos (\theta - \phi ) = \cos \theta \cos \phi + \sin \theta \sin \phi $ 

$ = \frac{3}{5}.\frac{4}{5} + \frac{4}{5}.\frac{3}{5} = \frac{{24}}{{25}}$ 

But $2{\cos ^2}\left( {\frac{{\theta - \phi }}{2}} \right) = 1 + \cos (\theta - \phi ) = 1 + \frac{{24}}{{25}}= \frac{{49}}{{50}}$ 

$\therefore$ ${\cos ^2}\left( {\frac{{\theta - \phi }}{2}} \right) = \frac{{49}}{{50}}$. 

Hence, $\cos \left( {\frac{{\theta - \varphi }}{2}} \right) = \frac{7}{{5\sqrt 2 }}$.

then, $\frac{{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\cos }^2}\frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = y$

==> $\frac{{2\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2} + \sin \frac{\alpha }{2}}} \times \frac{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}}{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}} = y$

==> $\frac{{1 - \cos \alpha + \sin \alpha }}{{1 + \sin \alpha }} = y$.

$\Rightarrow \tan \beta = \frac{1}{3}$

==> $\tan 2\beta = \frac{{2\tan \beta }}{{1 - {{\tan }^2}\beta }} = \frac{3}{4}$

$\therefore \tan (\alpha + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$

Now, $0 < \beta < \frac{\pi }{2}$ and $\tan 2\beta = \frac{3}{4} > 0$ both 

==> $0 < 2\beta < \frac{\pi }{2}$. 

Again,$0 < \alpha < \frac{\pi }{2}$ and $0 < 2\beta < \frac{\pi }{2}$ both 

==> $0 < \alpha + 2\beta < \pi $

Thus, $0 < \alpha + 2\beta < \pi $ and $\tan (\alpha + 2\beta ) = 1$ both 

==> $\alpha + 2\beta = \frac{\pi }{4} $

$\Rightarrow 2\beta = \frac{\pi }{4} - \alpha $.

$ = 16(2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A)$ 

$ = 16\left( {2 \times \frac{9}{{16}} - 1 - 4 \times \frac{{27}}{{64}} + 3 \times \frac{3}{4}} \right) = 11$.

$\sin 2\beta = \frac{{2T}}{{1 + {T^2}}} = \frac{3}{5} \Rightarrow \cos 2\beta = \frac{4}{5}$   {$T = \tan \beta $}

$\therefore \,\,\sin 4\beta = 2\sin 2\beta \cos 2\beta $

$= 2.\frac{3}{5}.\frac{4}{5} = \frac{{24}}{{25}} = \cos 2\alpha $.

$\sin 2\theta = {x^2} - 1 \le 1 \Rightarrow {x^2} \le 2$

or $ - \sqrt 2 \le x \le \sqrt 2 $          $[\because \,\,\sin 2\theta  \le 1]$

Now ${\sin ^6}\theta + {\cos ^6}\theta $

$ = {({\sin ^2}\theta + {\cos ^2}\theta )^3} - 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )$

$ = 1 - 3{\sin ^2}\theta {\cos ^2}\theta = 1 - \frac{3}{4}{\sin ^2}2\theta $

$ = 1 - \frac{3}{4}{({x^2} - 1)^2} = \frac{1}{4}\{ 4 - 3{({x^2} - 1)^2}\} $

Thus the given result will hold true only when ${x^2} \le 2$ and not for all real values of $x.$

$= \frac{{\sqrt 2 - \sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}{{\sqrt 2 \left\{ {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } \right\}}}$

$=\frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right)}}$

$= \frac{{\sqrt 2 \left\{ {\,1 - \cos \theta } \right\}}}{{\sqrt 2 \sin \theta }},$   where $\theta = \alpha - \frac{\pi }{4}$

$= \frac{{2{{\sin }^2}(\theta /2)}}{{2\sin (\theta /2)\cos (\theta /2)}} = \tan \frac{\theta }{2}$

$ = \tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$.

$\therefore$ $25{\cos ^2}\alpha + 5\cos \alpha - 12 = 0$ 

==> $\cos \alpha = \frac{{ - 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ - 5 \pm 35}}{{50}}$ 

==> $\cos \alpha = - 4/5$     $[ \because \pi /2 < \alpha  < \pi  \Rightarrow \cos \alpha  < 0]$ 

$\therefore $$\sin 2\alpha = 2\sin \alpha \cos \alpha = - 24/25$.

==> $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$

Hence, $\sqrt {1 + \sin A} = \sin \frac{A}{2} + \cos \frac{A}{2}$…..$(i)$

and $\sqrt {1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2}$…..$(ii)$

Subtract $(ii)$ from $(i)$, $2\cos \frac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} $.

==> $\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$,   [Let $\tan B = t$] 

==> $\sin 2B = \frac{{2t}}{{1 + {t^2}}},\cos 2B = \frac{{1 - {t^2}}}{{1 + {t^2}}}$ 

$\therefore$ $\frac{{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)}}{{5 - \left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right)}}$

$ = \frac{{2t}}{{4 + 6{t^2}}} = \frac{t}{{2 + 3{t^2}}} = \tan (A - B)$.

==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$

$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$ 

==> $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{1}{3}$.

and $\cos 2\,\theta + \cos 2\,\varphi = 3/2$…..$(ii)$ 

Square અને adding , 

$\therefore \,({\sin ^2}2\theta + {\cos ^2}2\theta ) + ({\sin ^2}2\phi + {\cos ^2}2\phi  )$

$ + 2\,[\sin 2\,\theta \,\sin 2\,\phi + \cos 2\,\theta \,\cos 2\,\phi ] = 1/4 + 9/4$ 

==> $\cos 2\theta \cos 2\,\phi + \sin 2\theta \sin 2\phi = 1/4$ 

==> $\cos (2\theta - 2\phi ) = 1/4$ 

==> ${\cos ^2}(\theta - \phi ) = 5/8$.

Now, put $\theta = \phi = \frac{\pi }{4}$

$\cos 2\left( {\frac{\pi }{2}} \right) - 4\cos \left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{4}} \right) + 2{\sin ^2}\left( {\frac{{2\pi }}{4}} \right) = 0$

Put $\theta = \phi = \pi /4$  in option $(a)$, 

then, $\cos 2\theta = \cos \pi /2 = 0$.

Hence option $(a)$ is correct.

$ = \frac{{\sin A - \cos A + 1}}{{\sin A - 1 + \cos A}} $

$= \frac{{\sin A + (1 - \cos A)}}{{\sin A - (1 - \cos A)}}$

$ = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2} + 2{{\sin }^2}\frac{A}{2}}}{{2\sin \frac{A}{2}\cos \frac{A}{2} - 2{{\sin }^2}\frac{A}{2}}}$

$ = \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}} $

$= \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}^2}}}{{{{\cos }^2}\frac{A}{2} - {{\sin }^2}\frac{A}{2}}}$

$ = \frac{{1 + \sin A}}{{\cos A}}$.

Then $N = 3\sin \theta - 4{\sin ^3}\theta - (4{\cos ^3}\theta - 3\cos \theta )$ 

$ = 3(\sin \theta + \cos \theta ) - 4({\sin ^3}\theta + {\cos ^3}\theta )$ 

$ = (\sin \theta + \cos \theta )\{ 3 - 4({\sin ^2}\theta - \sin \theta \cos \theta + {\cos ^2}\theta )\} $ 

$\therefore \;\frac{N}{D} + 1 $

$=  \frac{{(\sin \theta + \cos \theta )\{ 3 - 4(1 - \sin \theta \cos \theta )\} }}{{\sin \theta + \cos \theta }} + 1$

$ = 3 - 4(1 - \sin \theta \cos \theta ) + 1$

$ = 4\sin \theta \cos \theta = 2\sin 2\theta $.

$ = - 1 - 2\cos C\cos (A - B) + 2{\cos ^2}C$ 

$ = - 1 - 2\cos C[\cos (A - B) + \cos (A + B)]$

$ = - 1 - 4\cos A\cos B\cos C$.

and ${N^r} = 4\sin A\sin B\sin c$ 

$\therefore L.H.S. = \frac{{{N^r}}}{{{D^r}}}$

and $\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}$ .

$ = 1 - {\cos ^2}A + 1 - {\cos ^2}B + {\sin ^2}C$

$ = 2 - {\cos ^2}A - \cos (B + C)\cos (B - C)$

$ = 2 - \cos A[\cos A - \cos (B - C)]$

$ = 2 - \cos A[ - \cos (B + C) - \cos (B - C)]$ 

$ = 2 + \cos A.2\cos B\cos C$  

$\therefore$ ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\cos B\cos C = 2$.

$\therefore \tan A\tan B\tan C = 6 \Rightarrow \tan C = \frac{6}{2} = 3$ 

Also $\tan A + \tan B = 6 - 3 = 3$ 

$\therefore \tan A,\tan B = 2,1$ or $1, 2$ and $\tan C = 3$.

$= \frac{{{S_1} - {S_3}}}{{1 - {S_2}}} = \tan \frac{\pi }{2} = \infty $ 

$\therefore {S_2} = 1$ or $xy + yz + zx = 1$, 

where $x = \tan \frac{A}{2}$etc. 

Now ${(x - y)^2} + {(y - z)^2} + {(z - x)^2} \ge 0$

or $2\sum {x^2} - 2\sum xy \ge 0 \Rightarrow \sum {x^2} \ge 1$.   $\{ \because \sum xy = 1\} $

$A + B + C = \pi \Rightarrow B + C = \pi - A$ 

$\therefore \cos (B + C) = \cos (\pi - A) \Rightarrow \cos (B + C) = - \cos A$

$ \Rightarrow \cos B\cos C - \sin B\sin C = - \cos B\cos C$

$( \because {\rm{Given}}\cos A = \cos B\cos C)$ 

$ \Rightarrow 2\cos B\cos C = \sin B\sin C$ 

$ \Rightarrow \frac{{\cos B\cos C}}{{\sin B\sin C}} = \frac{1}{2}$

$\Rightarrow \cot B\cot C = \frac{1}{2}$.

$ = \frac{{\sin (B + C)}}{{\sin B\,\sin C}}$

$ = \frac{{\sin ({{180}^o} - A)}}{{\sin B\,\sin C}}$

$ = \frac{{\sin A}}{{\sin B\sin C}}$

Similarly, $\cot C + \cot A = \frac{{\sin B}}{{\sin C\sin A}}$ 

and $\cot A + \cot B = \frac{{\sin C}}{{\sin A\sin B}}$ 

Therefore, $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$

$ = \frac{{\sin A}}{{\sin B\sin C}}.\frac{{\sin B}}{{\sin C\sin A}}.\frac{{\sin C}}{{\sin A\sin B}}$

$ = \cos {\rm{ec}}A\cos {\rm{ec}}B\cos {\rm{ec}}C.$

$\therefore \,\frac{A}{2} + \frac{B}{2} = {90^o} - \frac{C}{2}$

$\therefore \cot \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \left( {{{90}^o} - \frac{C}{2}} \right)$

or $\frac{{\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\, - 1}}{{\cot \frac{B}{2}\, + \,\cot \frac{A}{2}}} = \tan \frac{C}{2} = \frac{1}{{\cot \frac{C}{2}}}$ 

or $\left( {\cot \frac{A}{2}\cot \frac{B}{2} - 1} \right)\cot \frac{C}{2} = \cot \frac{B}{2} + \cot \frac{A}{2}$ 

$\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\,.\,\cot \frac{C}{2} = \cot \frac{C}{2} + \cot \frac{B}{2}$ $ + \cot \frac{A}{2}.$

$ \Rightarrow \tan (A + B) = \tan (\pi - C)$ 

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan C}} = \tan (\pi - C)$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C$ 

Now $C$ is an obtuse angle, hence 

$ \Rightarrow \tan C < 0 \Rightarrow - \tan C > 0$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} > 0$

$\Rightarrow 1 - \tan A\tan B > 0$

$(\because A,B$ are acute angles; $\therefore \tan A > 0,\tan B > 0 )$

$ \Rightarrow \tan A\tan B < 1$.

$ = 2\cos (A + B)\cos (A - B) + \cos 2C$ 

$ = 2\cos \left( {\frac{{3\pi }}{2} - C} \right)\cos (A - B) + \cos 2C$

$ = - 2\sin C\cos (A - B) + 1 - 2{\sin ^2}C$

$ = 1 - 2\sin C\{ \cos (A - B) + \sin C\} $ 

$ = 1 - 2\sin C\left\{ {\cos (A - B) + \sin \left( {\frac{{3\pi }}{2} - (A + B)} \right)} \right\}$

$ = 1 - 2\sin C\{ \cos (A - B) - \cos (A + B)\} $

$ = 1 - 4\sin A\sin B\sin C$. 

Trick : Check by assuming $A = B = C = \frac{\pi }{2}$.

$ \Rightarrow \frac{{n - 1}}{{n + 1}} = \frac{{\sin A - \sin B}}{{\sin A + \sin B}} $

$= \frac{{2\cos \frac{{A + B}}{2}\sin \frac{{A - B}}{2}}}{{2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2}}}$ 

$ = \tan \frac{{A - B}}{2}\cot \frac{{A + B}}{2}$ 

$ \Rightarrow \frac{{n - 1}}{{n + 1}}\tan \left( {\frac{{A + B}}{2}} \right) = \tan \frac{{A - B}}{2}$ .

$ \Rightarrow \tan \theta = \frac{{\sin \left( {\alpha - \frac{\pi }{4}} \right)}}{{\cos \left( {\alpha - \frac{\pi }{4}} \right)}} $

$\Rightarrow \tan \theta = \tan \left( {\alpha - \frac{\pi }{4}} \right)$

$ \Rightarrow \theta = \alpha - \frac{\pi }{4}$

$\Rightarrow \alpha = \theta + \frac{\pi }{4}$

Hence, $\sin \alpha + \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) + \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \sqrt 2 \cos \theta $

and $\sin \alpha - \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) - \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \frac{1}{{\sqrt 2 }}\sin \theta + \frac{1}{{\sqrt 2 }}\cos \theta - \frac{1}{{\sqrt 2 }}\cos \theta + \frac{1}{{\sqrt 2 }}\sin \theta $

$ = \frac{2}{{\sqrt 2 }}\sin \theta = \sqrt 2 \sin \theta $.

using formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ and on solving,

we get the required result $i.e.$ $K = \frac{3}{4}$.

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ \,(2\sin {40^o}\sin 80^\circ )$

$ = \frac{1}{2}\sin 20^\circ \sin 60^\circ (\cos 40^\circ - \cos 120^\circ )$

$ = \frac{1}{2}.\frac{{\sqrt 3 }}{2}\sin 20^\circ \left( {1 - 2{{\sin }^2}20^\circ + \frac{1}{2}} \right)$

$ = \frac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\frac{3}{2} - 2{{\sin }^2}20^\circ } \right)$

$ = \frac{{\sqrt 3 }}{8}(3\sin 20^\circ - 4{\sin ^3}20^\circ )$

$ = \frac{{\sqrt 3 }}{8}\sin 60^\circ = \frac{{\sqrt 3 }}{8}.\frac{{\sqrt 3 }}{2} = \frac{3}{{16}}$.

==> $a\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = c$ 

$ \Rightarrow $ $a - a{\tan ^2}\theta + 2b\tan \theta = c + c{\tan ^2}\theta $

$ \Rightarrow $$ - (a + c){\tan ^2}\theta + 2b\,\tan \theta + (a - c) = 0$

$\therefore \tan \alpha + \tan \beta = - \frac{{2b}}{{ - (c + a)}} = \frac{{2b}}{{c + a}}$ .

==> $\frac{1}{{\sin \theta }} = \frac{{p + q}}{{p - q}}$

Apply componendo and dividendo 

$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{p + q + p - q}}{{p + q - p + q}}$

==> ${\left\{ {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$ 

==> ${\left\{ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$ 

==> ${\tan ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{p}{q}$

==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{q}{p}$

Note : $\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \sqrt {\frac{q}{p}} \,{\rm{only,}}\,\,{\rm{if}}$

$\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) > 0$.

$\frac{2}{{\cos 2\alpha }} = \frac{{\sin \beta }}{{\cos \beta }} + \frac{{\cos \beta }}{{\sin \beta }}$

$= \frac{{{{\sin }^2}\beta + {{\cos }^2}\beta }}{{\cos \beta \sin \beta }}$

$ = \frac{1}{{\cos \beta .\sin \beta }}$ 

==> $\cos 2\alpha = \sin 2\beta $ 

==> $\cos 2\alpha $= $\cos \,\left( {\frac{\pi }{2} - 2\beta } \right)$

==> $2\alpha = \frac{\pi }{2} - 2\beta $ 

==> $2\alpha + 2\beta = \frac{\pi }{2}$ 

==> $\alpha + \beta = \frac{\pi }{4}$.

==> $x = k\cos \theta $, $y = k\cos \left( {\theta - \frac{{2\pi }}{3}} \right)$,

$z = k\cos \left( {\theta + \frac{{2\pi }}{3}} \right)$ 

==> $x + y + z = k\left[ {\cos \theta + \cos \left( {\theta - \frac{{2\pi }}{3}} \right) + \cos \left( {\theta + \frac{{2\pi }}{3}} \right)} \right]$

$ = k[(0) = 0$ 

$ \Rightarrow $ $x + y + z = 0$.

$ = 2\,[3\sin \theta - 4{\sin ^3}\theta ]\,[4{\cos ^3}\theta - 3\cos \theta ]$ 

$=24 \sin \theta \cos \theta (\sin ^2 \theta + \cos ^2 \theta) -18\sin \theta \cos \theta  -32 \sin ^2 \theta \cos^2 \theta$

$ = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3\sin 2\theta $ 

On comparing, $x = \sin 2\theta .$

$\Rightarrow \sin A = - \frac{4}{5},\cos A = - \frac{3}{5}$

$\therefore $ $5\sin 2A + 3\sin A + 4\cos A$ 

$= 10\sin A\cos A + 3\sin A + 4\cos A$ 

$= 10\,\left( {\frac{{12}}{{25}}} \right) - \frac{{12}}{5} - \frac{{12}}{5} = 0$.

==>$A = {\cos ^2}\theta + {\sin ^2}\theta .{\sin ^2}\theta $

==> $A \le {\cos ^2}\theta + {\sin ^2}\theta $,     $[ \because {\sin ^2}\theta  \le 1]$

==> $A \le 1$

Again $A = {\cos ^2}\theta + {\sin ^4}\theta = (1 - {\sin ^2}\theta ) + {\sin ^4}\theta $

$A = {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}$

Hence, $3/4 \le A \le 1$.

$ = 2{\left( {\cos \theta + \frac{1}{2}} \right)^2} - \frac{3}{2}$

Now $2{\left( {\cos \theta + \frac{1}{2}} \right)^2} \ge 0$ for all $\theta $ 

$\therefore \,\,2{\left( {\cos \theta + \frac{1}{2}} \right)^2} - \frac{3}{2} \ge \frac{{ - 3}}{2}$ for all $\theta $.

==> $\cos 2\theta + 2\cos \theta \ge \frac{{ - 3}}{2}$ for all $\theta $ 

Also max. value of this expression is $3.$

==> $y = f(x) = {\cos ^2}x + {\sin ^2}x(1 - {\cos ^2}x)$

==> $y = {\cos ^2}x + {\sin ^2}x - {\sin ^2}x{\cos ^2}x$

==> $y = 1 - {\sin ^2}x{\cos ^2}x$

==> $y = 1 - \frac{1}{4}.{\sin ^2}2x$

$\therefore$ $\frac{3}{4} \le f(x) \le 1$,$(\because 0\le {{\sin }^{2}}2x\le 1)$

==> $f(R) \in [3/4,\,\,1]$.

$=2 \cos ^{2} \theta-1+\cos \theta$

$=-1+2\left(\cos ^{2} \theta+\frac{1}{2} \cos \theta+\frac{1}{16}\right)-\frac{1}{8}$

$=-\frac{9}{8}+2\left(\cos \theta+\frac{1}{4}\right)^{2} \geq-\frac{9}{8}$

So, the minimum value $S=-9 / 8$

==> $\tan (\alpha + \beta ) = \frac{{\frac{1}{{1 + \frac{1}{{{2^x}}}}} + \frac{1}{{1 + {2^{x + 1}}}}}}{{1 - \frac{1}{{1 + 1/{2^x}}}\frac{1}{{1 + {2^{x + 1}}}}}}$

==> $\tan (\alpha + \beta ) = \frac{{{2^x} + {{2.2}^{x + x}} + {2^x} + 1}}{{1 + {2^x} + {{2.2}^x} + {{2.2}^{x + x}} - {2^x}}}$

==> $\tan (\alpha + \beta ) = 1$

$ \Rightarrow \alpha + \beta = \frac{\pi }{4}$.

${x^2} + {y^2} = 1 + 1 + 2\,\cos \,(2A - A)$

$\therefore \,\,\,\frac{{{x^2} + {y^2} - 2}}{2} = \cos \,A$…..$(i)$

Also $\cos A + 2\,{\cos ^2}A - 1 = y$or $(\cos A + 1)\,(2\,\cos A - 1) = y$

Put for $\cos A$ from $(i)$ and get the answer.

$\therefore \,\,{\sin ^2}\beta = \sin \alpha \cos \alpha $

Now $\cos 2\beta = 1 - 2{\sin ^2}\beta = 1 - 2\sin \alpha \cos \alpha $

$ = {(\cos \alpha - \sin \alpha )^2} = 2\,{\left( {\frac{1}{{\sqrt 2 }}\cos \alpha - \frac{1}{{\sqrt 2 }}\sin \alpha } \right)^2}$

$ = 2{\sin ^2}\left( {\frac{\pi }{4} - \alpha } \right)$,

which is given in $(a).$

Also $\cos 2\beta = 2{\cos ^2}\left\{ {\frac{\pi }{2} - \left( {\frac{\pi }{4} - \alpha } \right)} \right\} $

$= 2{\cos ^2}\left( {\frac{\pi }{4} + \alpha } \right)$,

which is given in $(b).$

On squaring both sides,

we get $x + \frac{1}{x} + 2 = 4\,{\cos ^2}\theta $

==> $x + \frac{1}{x} = 4{\cos ^2}\theta - 2$

==> $x + \frac{1}{x} = $$2(2{\cos ^2}\theta - 1)$ $ = 2\cos 2\theta $…..$(ii)$

Again squaring both sides, ${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}2\theta $

==> ${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}2\theta - 2$$ = 2(2{\cos ^2}2\theta - 1)$

==> ${x^2} + \frac{1}{{{x^2}}} = 2\cos 4\theta $…..$(iii)$

Now take cube of both sides, ${\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^3} = {(2\cos 4\theta )^3}$

==> ${x^6} + \frac{1}{{{x^6}}} + 3{x^2} \times \frac{1}{{{x^2}}}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 8{\cos ^3}4\theta $

==> ${x^6} + \frac{1}{{{x^6}}} + 3\,(2\cos 4\theta ) = 8{\cos ^3}4\theta $

$ \Rightarrow {x^6} + \frac{1}{{{x^6}}} = 8{\cos ^3}4\theta - 6\cos 4\theta $

$= 2\,(4{\cos ^3}4\theta - 3\cos 4\theta )$

$= 2\cos 3(4\theta )  =  2\cos 12\theta $.

==> $\frac{A}{2} = \frac{\pi }{2} - \left( {\frac{{B + C}}{2}} \right)$  

$\therefore$ $\cot \frac{A}{2} = \tan \left( {\frac{B}{2} + \frac{C}{2}} \right)$ 

==> $\frac{1}{{\tan \frac{A}{2}}} = \frac{{\tan \frac{B}{2} + \tan \frac{C}{2}}}{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}$

==> $1 - \tan \frac{B}{2}\tan \frac{C}{2} $

$= \tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{A}{2}.\tan \frac{C}{2}$

$\tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{A}{2}\tan \frac{C}{2} = 1$ 

$i.e.$, $\sum {\tan \frac{A}{2}\tan \frac{B}{2} = 1} $.

If $n$ is even, answer is $(b)$ and if $n$ is odd answer is $(a).$

$\therefore \,\,\,{(96\sin \alpha )^2} - (4 \times 64 \times 27) \ge 0$
==> ${\sin ^2}\alpha \ge \frac{{4 \times 64 \times 27}}{{96 \times 96}} \Rightarrow \sin \alpha \ge \pm \sqrt 3 /2$
For $\sin \alpha \ge \sqrt 3 /2,\alpha \in (\pi /3,\,2\pi /3)$
For $\sin \alpha \ge - \sqrt 3 /2,\alpha \in (4\pi /3,\,5\pi /3)$.

$P=\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$

Assume $x \leq y \leq z$

for $P$ to be max take $x=0, z=1$

$P =\sqrt{| y |}+\sqrt{| y -1|}+1$

$P =\sqrt{ y }+\sqrt{1- y }+1$

Put $y=\sin ^2 \theta$

$P =\sin \theta+\cos \theta+1$

$P _{\max }=\sqrt{2}+1$

We have,

$x(\theta)=|\cos 4 \theta| \cos \theta$ and $y(\theta)=|\cos 4 \theta| \sin \theta$

Let $S=\sum \limits_{k=0}^{44} \frac{1}{\cos k^{\circ} \cos (k+1)^{\circ}}$

$\Rightarrow \quad S=\frac{1}{\sin 1^{\circ}} \sum \limits_{k=0}^{44} \frac{\sin (k+1-k)^{\circ}}{\cos k^{\bullet} \cos (k+1)^{\circ}}$

$S=\frac{1}{\sin 1^{\circ}}$

$S=\frac{1}{\sin 1^{\circ}} \sum \limits_{k=0}^{44}\left(\tan (k+1)^{\circ}-\tan k^{\circ}\right)$

$S=\frac{1}{\sin 1^{\circ}}\left[\tan 1^{\circ}-\tan 0^{\circ}+\tan 2^{\circ}\right.$

$S=\frac{1}{\sin 1^{\circ}} \tan 45^{\circ}$

$S=\frac{1}{\sin 1^{\circ}}=\frac{1}{0.0174}=57.29$

$\left[\because \sin 1^{\circ}=0.017\right]$

$[S]=[57.29]=57$

Given, $C(\theta)=\sum_{n=0}^{\infty} \frac{\cos (n \theta)}{n !}$

$C(0)=\sum_{n=0}^{\infty} \frac{1}{n !}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots=e$

$C(\pi)=\sum_{n=0}^{\infty}(-1)^n \frac{1}{n !} \quad\left[\because \cos n \pi=(-1)^n\right]$

$C(\pi)=1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots=e^{-1}$

(A) $C(0), C(\pi)=e \cdot e^{-1}=1$ T'rue

(B) $C(0)+C(\pi)=e+\frac{1}{e}>2$ True

(C) $C(\theta)>0 \forall \theta \in R$ True

(D) $C^{\prime}(\theta)=\sum_{n=0}^{\infty}-\frac{n \sin (n \theta)}{n !}$

$\therefore \quad C^{\prime}(\theta)=0 \Rightarrow \theta=0$ False

Hence, option $(d)$ is false.

At 6: 15,

the minutes hand makes an angle is $\alpha$.

$\therefore \quad \alpha =90^{\circ}+15 \times\left(\frac{1}{2}\right)^{\circ}$

$\alpha =\left(\frac{195}{2}\right)^{\circ}$

and hour hand is $\beta$.

Given, $\alpha+\beta=360^{\circ}$

$\therefore \quad \beta =360^{\circ}-\alpha$

$=360^{\circ}-\left(\frac{195}{2}\right)^{\circ}=\left(\frac{525}{2}\right)^{\circ}$

Difference between their angles

$=\frac{525}{2}-\frac{195}{2}=\frac{390}{2}=165^{\circ}$

We have,

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$

$=\left(\tan 81^{\circ}+\tan 9^{\circ}\right)-\left(\tan 63^{\circ}+\tan 27^{\circ}\right)$

$=\left(\cot 9^{\circ}+\tan 9^{\circ}\right)-\left(\cot 27^{\circ}+\tan 27^{\circ}\right)$

$=\left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right)-\left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}}+\frac{\sin 27^{\circ}}{\cos 27^{\circ}}\right)$

$=\left(\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}\right)-\left(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}\right)$

$=\left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}\right)-\left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}\right)$

$=\left[\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}\right]$

$=2\left[\frac{1}{\sin 18^{\circ}}-\frac{1}{\cos 36^{\circ}}\right]=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right]$

$\left[\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}, \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right]$

$=8\left[\frac{\sqrt{5}+1-\sqrt{5}+1]}{5-1}\right]=4$

We have,

$\begin{array}{l}\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right)\left(1+\tan 3^{\circ}\right) \\ \text { We know that, } \\ (1+\tan \theta)\left(1+\tan \left(45^{\circ}-\theta\right)\right)=2 \\ \therefore\left(1+\tan 45^{\circ}\right) \\ \left(1+\tan 43^{\circ}\right)\left(1+\tan 44^{\circ}\right)\left(1+\tan 2^{\circ}\right) \\ \left(1+\tan 45^{\circ}\right) \\ \Rightarrow \quad 2^{22} \cdot 2=2^{23} \end{array}$

Exactly at $10\,O'clock$ the hour hand has travelled $300^{\circ}$ from $120^{\prime}$ 'clock.

One hour $=60$ minute.

One minute hand moves $1^{\circ}$ and hour clock hand move $\left(\frac{30}{360}\right)^{\circ}=\left(\frac{1}{12}\right)^{\circ}$

Assuming we have made it to 10 O'clock and now the hour and the minute hand start moving spontaneously.

If the hands of the watch are symmetric with vertical line.

Supposing this happens when $x$ minutes have passed $x$ minutes $=(6 x)^{\circ}$ have been covered our hour hand would cover.

On subtracting this from $360^{\circ}$ to find the angle from 12 O'clock anti-clockwise, we get

$360^{\circ}-\left(300+\frac{x}{2}\right)^{\circ}=\left(60-\frac{x}{2}\right)^{\circ}$

So, they are symmetric.

$\therefore \quad 60-\frac{x}{2}=6 x$

$\Rightarrow \quad x=\left(\frac{120}{13}\right)^{\circ}=9min\,13.8s$

$\therefore \text { Time }=10 h\,9m\,14s$

We have,

$\sin x+\sin y=\frac{7}{5}$

and $\quad \cos x+\cos y=\frac{1}{5}$

$\Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{7}{5}$

and $2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{1}{5}$

On dividing Eq.$(ii)$ by E.$(i)$, we get

$\tan \left(\frac{x+y}{2}\right)=7$

$\Rightarrow \quad \sin (x+y) =\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)}$

$\Rightarrow \quad \sin (x+y)=\frac{14}{1+49}=\frac{14}{50}$

$\Rightarrow \quad \sin (x+y)=\frac{7}{25}$

$ \tan x=\frac{3}{4} \cos x=-\frac{4}{5} $

$ 80\left(\tan ^2 x-\cos x\right) $

$ =80\left(\frac{9}{16}+\frac{4}{5}\right)=45+64=109$

$ \mathrm{x}-2 \geq 0 \& 4-\mathrm{x} \geq 0 $

$ \therefore \mathrm{x} \in[2,4] $

$ \text { Let } \mathrm{x}=2 \sin ^2 \theta+4 \cos ^2 \theta $

$ \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| $

$ \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} $

$ \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} $

$ \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} $

$ \alpha^2+2 \beta^2=2+40=42$

$\cos \theta \cos  ( 6 0^{ \circ } - \theta ) \cos (60^{\circ}+ \theta)=\frac{1}{4} \cos 3 \theta.$

So equation reduces to $\left|\frac{1}{4} \cos 3 \theta\right| \leq \frac{1}{8}$

$ \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} $

$ \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2}$

$\Rightarrow$ maximum value of $\cos 3 \theta=\frac{1}{2}$, here

$ \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} $

$ \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}$

As $\theta \in[0,2 \pi]$ possible values are

$\theta=\left\{\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}\right\}$

Whose sum is

$\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi$

$ a(\sin x-2)=2(\sin x-2)(\sin x+2) $

$ \sin x=2, \quad a=2(\sin x+2) $

$ \Rightarrow a \in[2,6] $

$ p=2 \quad q=6$

$ r=\tan 9^{\circ}+\cot 9^{\circ}-\tan 27-\cot 27 $

$ r=\frac{1}{\sin 9 \cdot \cos 9}-\frac{1}{\sin 27 \cdot \cos 27} $

$=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right] $

$ r=4 $

$ \text { p.q. } r=2 \times 6 \times 4=48$

$ \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 $

$ \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 $

$ \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 $

$ x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0$

so, sum of real solutions $=-1$

$ 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 $

$ 6 \sin x-4=0 $

$ \sin x=\frac{2}{3} $

${n}=5 \text { (in the given interval) } $

$ x^2+5 x+2=0$

$ x=\frac{-5 \pm \sqrt{17}}{2}$

$ \text { Required interval }(-\infty, 0)$

Required interval $(-\infty, 0)$

let $\tan \frac{\theta}{2}=\mathrm{t}$

$ \frac{4-4 t^2-6 t}{1+t^2}=1 $

$ 4-4 t^2-6 t=1+t^2$

$ \Rightarrow 5 t^2+6 t-3=0 $

$ \Rightarrow t=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)} $

$ =\frac{-6 \pm \sqrt{96}}{10} $

$ =\frac{-6 \pm 4 \sqrt{6}}{10} $

$ t=\frac{-3+2 \sqrt{6}}{5} $

$ \cos \theta=\frac{1-t^2}{1+t^2}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^2}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^2}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} $

$ =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}} $

$ =\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}} $

$ =\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}$

$ =\frac{4 \sqrt{5}-1}{\sqrt{5}+4} \times \frac{\sqrt{5}-4}{\sqrt{5}-4}$

$ =\frac{20-16 \sqrt{5}-\sqrt{5}+4}{-11} $

$ =\frac{17 \sqrt{5}-24}{11} \Rightarrow a=17, b=27, c=11 $

$ a+b+c=52$

$\Rightarrow \tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}$

$\Rightarrow \frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}$

$\Rightarrow \frac{2 \times 4}{\sqrt{5}-1}-\frac{2 \times 4}{\sqrt{5+1}}$

$\Rightarrow 4$

$2 P \times \sin \frac{\pi}{33}=96 \times 2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$

$2 P \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \sin \frac{\pi}{33}$

$P=3$

$\frac{1}{\tan 75^{\circ}}=\cot 75^{\circ}=2-\sqrt{3}$

$\frac{1}{\tan 105^{\circ}}=\cot \left(105^{\circ}\right)=-\cot 75^{\circ}=\sqrt{3}-2$

$\tan 195^{\circ}=\tan 15^{\circ}=2-\sqrt{3}$

$2(2-\sqrt{3})=2 a \Rightarrow a =2-\sqrt{3}$

$a +\frac{1}{ a }=4$

so given expression can be written as

$36 \times \frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}$

$36 \times \frac{\sin 729^{\circ}}{\sin 9^{\circ}}=36$

$Ar (\triangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 \tan \theta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(\tan \theta_1+\tan \theta_2\right)$

$BE = AB \left(\tan \theta_1+\cot \theta_1\right)$

$\frac{4}{\sqrt{3}} \tan \theta_1+\cot \theta_1 \Rightarrow \tan \theta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$

$\theta_1=\frac{\pi}{6}$

$\theta_1=\frac{\pi}{3} \quad \theta_2=\frac{\pi}{3}$

$as \frac{\theta_2}{\theta_1} \text { is } \operatorname{largest} \therefore \theta_1=\frac{\pi}{6} \theta_2=\frac{\pi}{3}$

$\therefore x ^2=\frac{(2 \sqrt{3}-3) \times 2}{\tan \theta_1}=\frac{\sqrt{3}(2-\sqrt{3}) \times 2}{\tan \frac{\pi}{6}}$

$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$

$x =3-\sqrt{3}$

Perimeter of $\triangle C E D$

$= CD + DE + CE$

$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) \times 2=6$

$=\sin 12^{\circ}-2 \cos 42^{\circ} \sin 30^{\circ}$

$=\sin 12^{\circ}-\sin 48^{\circ}$

$=-2 \cos 30^{\circ} \sin 18^{\circ}$

$=-2 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{5}-1}{4}$

$=\frac{\sqrt{3}}{4}(1-\sqrt{5})$

$=16 \sin 40^{\circ} \sin 20^{\circ} \sin 80^{\circ}$

$=4(4 \sin (60-20) \sin (20) \sin (60+20))$

$=4 \times \sin \left(3 \times 20^{\circ}\right)$

${[\because \sin 3 \theta=4 \sin (60-\theta) \times \sin \theta \times \sin (60+\theta)]}$

$=4 \times \sin 60^{\circ}$

$=4 \times \frac{\sqrt{3}}{2}=2 \sqrt{3}$

$=\frac{\sin \left(3 \times \frac{\pi}{7}\right)}{\sin \frac{\pi}{7}} \times \cos \left(\frac{\frac{2 \pi}{7}+\frac{6 \pi}{7}}{2}\right)$

$=\frac{2 \sin \left(\frac{3 \pi}{7}\right)}{2 \sin \frac{\pi}{7}} \times \cos \left(\frac{4 \pi}{7}\right)$

$=\frac{\sin \left(\frac{7 \pi}{7}\right)+\sin \left(\frac{-\pi}{7}\right)}{2 \sin \frac{\pi}{7}}$

$=\frac{-\sin \frac{\pi}{7}}{2 \sin \frac{\pi}{7}}$

$=-\frac{1}{2}$

$=2 \sin \left(\frac{11 \pi-10 \pi}{22}\right) \sin \left(\frac{11 \pi-8 \pi}{22}\right) \sin \left(\frac{11 \pi-6 \pi}{22}\right)$ $\sin \left(\frac{11 \pi-4 \pi}{22}\right) \sin \left(\frac{11 \pi-2 \pi}{22}\right)$

$=2 \cos \frac{\pi}{11} \cos \frac{2 \pi}{11} \cos \frac{3 \pi}{11} \cos \frac{4 \pi}{11} \cos \frac{5 \pi}{11}$

$=\frac{2 \sin \frac{32 \pi}{11}}{2^{5} \sin \frac{\pi}{11}}$

$=\frac{1}{16}$

$\tan (\alpha+\beta)=\frac{1-\frac{4}{3}}{1+\frac{4}{3} \times 1}=\frac{-1}{7}$

$\sin 10^{\circ} \frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \cdot \frac{1}{4} \sin 30^{\circ}$

$\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin 10^{\circ}\left(\cos 20^{\circ}-\frac{1}{2}\right)$

$=\frac{1}{32}\left(2 \sin 10^{\circ} \cos 20^{\circ}-\sin 10^{\circ}\right)$

$=\frac{1}{32}\left(\sin 30^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}\right)$

$=\frac{1}{32}\left(\frac{1}{2}-2 \sin 10^{\circ}\right)$

$=\frac{1}{64}\left(1-4 \sin 10^{\circ}\right)$

$=\frac{1}{64}-\frac{1}{16} \sin 10^{\circ}$

Hence $\alpha=\frac{1}{64}$

$16+\alpha^{-1}=80$

$2 \sin ^{2} \frac{\pi}{8} \sin ^{2} \frac{2 \pi}{8} \sin ^{2} \frac{3 \pi}{8}$

$\sin ^{2} \frac{\pi}{8} \sin ^{2} \frac{3 \pi}{8}$

$\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}$

$\frac{1}{4} \sin ^{2}\left(\frac{\pi}{4}\right)=\frac{1}{8}$

$16\left[4 \sin 2 \theta \cos ^{2} 2 \theta+2 \cos ^{2} 2 \theta-1\right]$

Now:

$\sin \theta+\cos \theta=\frac{1}{2}$

$1+\sin 2 \theta=\frac{1}{4}$

$\sin 2 \theta=-\frac{3}{4}$

$\cos ^{2} 2 \theta=1-\frac{9}{16}=\frac{7}{16}$

$16\left[-4(-3 / 4) \times \frac{7}{16}+2 \times \frac{7}{16}-1\right]$

$16\left[\frac{-7}{16}-1\right] \Rightarrow-23$

$15 \sin ^{4} \alpha+10 \cos ^{4} \alpha=6\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}$

$\left(3 \sin ^{2} \alpha-2 \cos ^{2} \alpha\right)^{2}=0$

$\tan ^{2} \alpha=\frac{2}{3} \cdot \cot ^{2} \alpha=\frac{3}{2}$

$\Rightarrow 27 \sec ^{6} \alpha+8 \operatorname{cosec}^{6} \alpha$

$=27\left(\sec ^{6} \alpha\right)^{3}+8\left(\operatorname{cosec}^{6} \alpha\right)^{3}$

$=27\left(1+\tan ^{2} \alpha\right) 3+8\left(1+\cot ^{2} \alpha\right)^{3}$

$=250$

$\Rightarrow k +1 \in\left[-\sqrt{3^{2}+4^{2}}, \sqrt{3^{2}+4^{2}}\right]$

$\Rightarrow k +1 \in[-5,5]$

$\Rightarrow k \in[-6,4]$

No. of integral values of $k =11$

put, $\theta=\frac{\pi}{24}$

$\left\{\therefore \cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \, \& \, \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\right\}$

$\Rightarrow \cot \left(\frac{\pi}{24}\right)=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)}$

$=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$

$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$

$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$

$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}\right)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow\left(\cos \left(\frac{x+y}{2}\right)-\frac{1}{2} \cos \left(\frac{x-y}{2}\right)\right)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and $\cos \left(\frac{x+y}{2}\right)=\frac{1}{2} \cos \left(\frac{x-y}{2}\right)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$\therefore \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

$\log _{10} \sin x+\log _{10} \cos x=-1$

$\Rightarrow \quad \log _{10} \sin x \cdot \cos x=-1$

$\Rightarrow \quad \sin x \cdot \cos x=\frac{1}{10}$ $....(1)$

$\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right)$

$\Rightarrow \quad \sin x+\cos x=10^{\left(\log _{10} \sqrt{n}-\frac{1}{2}\right)}=\sqrt{\frac{n}{10}}$

by squaring

$1+2 \sin x \cdot \cos x=\frac{n}{10}$

$\Rightarrow \quad 1+\frac{1}{5}=\frac{ n }{10} \quad \Rightarrow \quad n =12$

$\Rightarrow$ Centroid also lies on $y-$axis

$\Rightarrow \Sigma \cos \alpha=0$

$\quad \cos \alpha+\cos \beta+\cos \gamma=0$

$\Rightarrow \cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma=3 \cos \alpha \cos \beta \cos \gamma$

$\therefore \frac{\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma}$

$\frac{4\left(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma\right)-3(\cos \alpha+\cos \beta+\cos \gamma)}{\cos \alpha \cos \beta \cos \gamma}$

$=12$

Also, $\cos ^{2} \theta=\frac{1}{y} \& 1-\sin ^{2} \theta \cos ^{2} \theta=\frac{1}{z}$

So, $1-\frac{1}{x} \times \frac{1}{y}=\frac{1}{z} \Rightarrow z(x y-1)=x y \quad \ldots(1)$

Also, $\frac{1}{x}+\frac{1}{y}=1 \quad \Rightarrow x+y=x y$ $....(2)$

From $(i)$ and $(ii)$

$x y+z=x y z=(x+y) z$

Let $\operatorname{cosec} 18^{\circ}=\mathrm{x}=\sqrt{5}+1$

$\Rightarrow \mathrm{x}-1=\sqrt{5}$

Squaring both sides, we get

$x^{2}-2 x+1=5$

$\Rightarrow x^{2}-2 x-4=0$

$\sin \beta=\frac{1}{\sqrt{10}} \Rightarrow \tan \beta=\frac{1}{3} \Rightarrow \tan 2 \beta=\frac{3}{4}$

$\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}=1$

$=\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}=\frac{1}{2} \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$

$\tan \alpha . \tan \beta=\frac{\mathrm{k}-1}{\mathrm{k}+1}$

$\tan (\alpha+\beta)=\frac{\frac{\lambda \sqrt{2}}{\mathrm{k}+1}}{1-\frac{\mathrm{k}-1}{\mathrm{k}+1}}=\frac{\lambda \sqrt{2}}{2}=\frac{\lambda}{\sqrt{2}}$

$\Rightarrow \frac{\lambda^{2}}{2}=50 \Rightarrow \lambda=10 \;and\;-10$

$ = 3\,(1 - 2\sin \,2\theta  + {\sin ^2}2\theta ) + \,6 + 6\sin 2\theta  + \,4\,{\sin ^6}\theta $

$ = \,9 + 3{\sin ^2}2\theta  + 4\,{\sin ^6}\theta $

$ = \,9 + 12{\sin ^2}\theta {\cos ^2}\theta  + 4\,{(1 - {\cos ^2}\theta )^3}$

$ = \,9 + 12(1 - {\cos ^2}\theta ){\cos ^2}\theta  + 4\,(1 - 3{\cos ^2}\theta  + 3{\cos ^4}\theta  - {\cos ^6}\theta )$

$ = \,13 + 12{\cos ^2}\theta  - 12{\cos ^4}\theta  - 12{\cos ^2}\theta  + 12{\cos ^4}\theta  - 4{\cos ^6}\theta $

$ = \,13 - 4{\cos ^6}\theta $

Let $\,{\cos ^2}2\theta \, =t $

$ \Rightarrow \,1\, - \,\,{\cos ^2}2\theta \, + \,{\cos ^4}2\theta \, = \frac{3}{4}$

$ \Rightarrow t = \frac{1}{2}\, \Rightarrow \,\,{\cos ^2}2\theta \, = \frac{1}{2}\,$

$ \Rightarrow 2\,{\cos ^2}2\theta  - 1 = 0\, \Rightarrow \,\cos 4\theta \, = 0$

$ \Rightarrow \,4\theta \, = (2n + 1)\frac{\pi }{2}$

$ \Rightarrow \,\theta \, = (2n + 1)\frac{\pi }{8} \Rightarrow \theta  = \frac{\pi }{8},\frac{{3\pi }}{8} \in \left[ {0,\frac{\pi }{2}} \right]$

Sum of values of $\theta $ is $\frac {\pi }{2}$

$ \Rightarrow \frac{1}{2}\,(1 + \,\cos \,{20^o} - (\cos \,{60^o} + \cos \,{40^o})\, + 1 + \,\,\cos {100^o})$

$ \Rightarrow \frac{1}{2}\,\left( {\frac{3}{2} + \,\cos \,{{20}^o} + 2\sin \,{{70}^o}\sin \,( - {{30}^o})} \right)$

$ \Rightarrow \frac{1}{2}\,\left( {\frac{3}{2} + \,\cos \,{{20}^o} - \sin \,{{70}^o}} \right)$

$ \Rightarrow \frac{3}{4}$

$ = \,\,\sin \,{10^o}\,\sin \,{30^o}\,\sin \,{50^o}\,\sin \,{70^o}$

$ = \,\,\sin \,{30^o}\,\{ \sin \,{10^o}\sin \,({60^o} - {10^o})\,\sin ({60^o} + {10^o})\,\} $

$ = \,\,\sin \,{30^o}\,\left\{ {\frac{1}{4}\sin \,3({{10}^o})} \right\}$

$ = \,\frac{1}{2}\left( {\frac{1}{4} \times \frac{1}{2}} \right)$

$ = \,\frac{1}{{16}}$

$ = \,5\left( {\sin \,\theta \,\cos \frac{\pi }{6}\, - \,\cos \,\theta \sin \frac{\pi }{6}} \right)\, + \,3\cos \,\theta $

$ = \frac{{5\sqrt 3 }}{2}\sin \,\theta \, + \,\frac{1}{2}\cos \,\theta $

Maximum value is  $\sqrt {{{\left( {\frac{{5\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} \, = \,\sqrt {\frac{{76}}{4}} \, = \,\sqrt {19} $

${F_6}(x)\, = \,\frac{{{{\sin }^6}x + {{\cos }^6}x}}{6} = $ $\frac{{1 - 3{{\sin }^2}x.{{\cos }^2}x\,({{\sin }^2}x\, + \,{{\cos }^2}x)}}{6}$

$ = \frac{1}{6} - \frac{1}{2}{\sin ^2}x.{\cos ^2}x$

${F_4}\left( x \right) - {F_6}(x) = \frac{1}{4} - \frac{1}{6} = \frac{{6 - 4}}{{24}} = \frac{2}{{24}} = \frac{1}{{12}}$

If $\cos \,(\,\alpha  + \,\beta )\, = \,\frac{3}{5}$ then $\tan \,(\,\alpha  + \,\beta )\, = \,\frac{3}{4}$ and if $\sin \,(\,\alpha  - \,\beta )\, = \,\frac{5}{{13}}$ then $\tan \,(\,\alpha  - \,\beta )\, = \,\frac{5}{{12}}$

(since $\alpha  - \,\beta $ here lies in the first quadrant)

Now $\tan \,(\,2\alpha ) = \tan \{ (\alpha \, + \,\beta ) + (\alpha \, - \,\beta )\} $

$ = \,\frac{{\tan \,(\alpha \, + \,\beta ) + \tan \,(\alpha \, - \,\beta )}}{{1 - \tan \,(\alpha \, + \,\beta ).\tan \,(\alpha \, - \,\beta )}}$ 

$ = \frac{{\frac{4}{3} + \frac{5}{{12}}}}{{1 - \frac{4}{3}.\frac{5}{{12}}}} = \frac{{63}}{{16}}$

$\tan A+\tan B=\frac{10}{3}$

$\tan A \times \tan B=\frac{-25}{3}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\frac{\frac{10}{3}}{1+\frac{25}{3}}$

$\frac{\frac{10}{3}}{\frac{28}{3}}=\frac{10}{28}=\frac{5}{14}$

$\therefore \tan (A+B)=\frac{5}{14}$

$\therefore \sin (A+B)=\frac{5}{\sqrt{221}}$

$\cos (A+B)=\frac{14}{\sqrt{221}}$

$\therefore 3 \sin ^2(A+B)-10 \sin (A+B) \cos (A+B)-25 \cos ^2(A+B)$

$=3 \times \frac{25}{221}-10 \times \frac{70}{221}-25 \times \frac{196}{221}$

$=\frac{75-700-4900}{221}$

$=-25$

$ = \sqrt {1 + \sin {{240}^o}}  - \sqrt {1 - \sin {{240}^o}} $ $ = \sqrt {\frac{{1 - \sqrt 3 }}{2}}  - \sqrt {\frac{{1 + \sqrt 3 }}{2}}  \ne \sqrt 3 $

For statement $q$: $\frac{{A + C}}{2} + \frac{{B + D}}{2} = \pi $$ \Rightarrow \cos (\frac{{A + C}}{2}) + \cos (\frac{{B + D}}{2}) = 0$

So statement $p$ is False and statement $q$ is True

$5\,{\tan ^2}x\, - 5{\cos ^2}x = 2(2{\cos ^2}x - 1) + 9$

$ \Rightarrow \,5\,{\tan ^2}x\, - 5{\cos ^2}x = 4{\cos ^2}x - 2 + 9$

$ \Rightarrow \,5\,{\tan ^2}x = 9{\cos ^2}x + 7$

$ \Rightarrow \,5\,({\sec ^2}x - 1) = 9{\cos ^2}x + 7$

Let ${\cos ^2}x = t$

$ \Rightarrow \frac{5}{t} - 9t - 12 = 0$

$ \Rightarrow 9{t^2} + 12t - 5 = 0$

$ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0$

$ \Rightarrow (3t - 1)(3t + 5) = 0$

$ \Rightarrow t\, = \frac{1}{3}$ as $t\, \ne  - \frac{5}{3}$

$\cos \,2x = 2{\cos ^2}x - 1 = 2\left( {\frac{1}{3}} \right) - 1 =  - \frac{1}{3}$

$\cos \,4x = 2{\cos ^2}2x - 1 = 2{\left( { - \frac{1}{3}} \right)^2} - 1 =  - \frac{7}{9}$

$4\, + \,2(1 - {\cos ^2}x){\cos ^2}x - 2{\cos ^4}x$

$ - 4\left\{ {{{\cos }^4}x - \frac{{{{\cos }^2}x}}{2} - 1 + \frac{1}{{16}} - \frac{1}{{16}}} \right\}\, - \,4\left\{ {{{\left( {{{\cos }^2}x - \frac{1}{4}} \right)}^2} - \frac{{17}}{{16}}} \right\}$

$0\, \leqslant \,{\cos ^2}x\, \leqslant 1$

$ - \frac{1}{4}\, \leqslant \,\,{\cos ^2}x\, - \,\frac{1}{4}\, \leqslant \,\frac{3}{4}$

$0\, \leqslant \,{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)^2}\, \leqslant \,\frac{9}{{16}}$

$ - \frac{{17}}{{16}}\, \leqslant \,{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)^2}\, - \frac{{17}}{{16}}\, \leqslant \,\frac{9}{{16}}\, \leqslant \, - \frac{{17}}{{16}}\,$

$\frac{{17}}{4} \geqslant  - \,4\left\{ {\,{{\left( {\,{{\cos }^2}x - \frac{1}{4}} \right)}^2} - \frac{{17}}{{16}}\,\,} \right\}\, \geqslant \,\frac{1}{2}$

$M = \frac{{17}}{4}$

$m = \frac{1}{2}$

$M - m\, = \,\frac{{17}}{4} - \frac{2}{4}\, = \frac{{15}}{4}$

$ \Rightarrow \,\frac{1}{{\sqrt 3 }}\, = \,\frac{y}{{1 - \tan \,A\,\tan \,B}}$ where $y\, = \tan \,A\, + \tan \,B$

$ \Rightarrow \tan \,A\,\tan \,B\, = \,1 - \sqrt 3 y$ Also $AM \geqslant GM$

$ \Rightarrow \,\frac{{\tan \,A\, + \,\tan \,B}}{2}\, \geqslant \,\sqrt {\tan \,A\,\tan \,B} $

$ \Rightarrow \,y\, \geqslant \,2\sqrt {1 - \sqrt 3 y} $

$ \Rightarrow \,{y^2}\, \geqslant \,4 - 4\sqrt 3 y$

$ \Rightarrow \,{y^2}\, + \,4\sqrt 3 y - 4 \geqslant 0$

$ \Rightarrow \,y\, \leqslant \, - \,2\sqrt 3  - 4$ or $ \Rightarrow \,y\,  \geqslant  \, - \,2\sqrt 3  + 4$

( $y\, \leqslant \, - \,2\sqrt 3  - 4$ is not possible as $\tan A\,\tan B\, > 0$ 

$ = \frac{1}{4}(1 - 2{\sin ^2}{\cos ^2}x) - \frac{1}{6}(1 - 3{\sin ^2}{\cos ^2}x)$

$ = \frac{1}{4} - \frac{1}{6} = \frac{{3 - 2}}{{12}} = \frac{1}{{12}}$

$f(x)\, = \,3\,\sin x + 4\,\cos x$

Now,

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left( {\frac{{ - 11\pi }}{6}} \right)\, + 4\,\cos \left( {\frac{{ - 11\pi }}{6}} \right)\,$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left( { - 2\pi  + \frac{\pi }{6}} \right)\, + 4\,\cos \left( { - 2\pi  + \frac{\pi }{6}} \right)\,$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \,3\,\sin \left\{ { - \left( {2\pi  - \frac{\pi }{6}} \right)} \right\}\, + 4\,\cos \left\{ { - \left( {2\pi  - \frac{\pi }{6}} \right)} \right\}$

$\left\{ {For\,\,odd\,\,functions\,\begin{array}{*{20}{c}}
  {\sin \theta \, = \, - \,\sin \theta } \\ 
  {and\,\,\cos \,( - \theta ) =  - \cos \,\,\theta } 
\end{array}} \right\}$

$f\left( {\frac{{ - 11\pi }}{6}} \right)\, = \, - 3\,\sin \left( {2\pi  - \frac{\pi }{6}} \right)\, - 4\,\cos \left( {2\pi  - \frac{\pi }{6}} \right)$

$ \Rightarrow \,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\, + 3\sin \left( {\frac{\pi }{6}} \right) - 4\,\cos \left( {\frac{\pi }{6}} \right)$

$ \Rightarrow \,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\,3 \times \frac{1}{2} - 4\, \times \frac{{\sqrt 3 }}{2}$

or $\,f\left( {\frac{{ - 11\,\pi }}{6}} \right)\, = \,\,\frac{3}{2} - 2\sqrt 3 $

$ = \frac{1}{{\sin A - \cos A}}\left\{ {\frac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A}}} \right\}$

$ = \frac{{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}A}}{{\sin A\cos A}}$ $ = 1 + \sec A\cos ecA$

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$

$=\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x\right\}$

$=\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right)$

$=\cos \frac{x}{2}+\sin \frac{x}{2}$

$=\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}}$

$=\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text { Option (B) is correct. }$

$\cot x=-\frac{5}{\sqrt{11}}$

$\frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11}$, As $\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

$\Rightarrow \frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}$

$\Rightarrow x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4} ; n \in Z$

$\Rightarrow x=2 n \pi ; x=2 n \pi+\frac{\pi}{2} ; n \in Z$

$\Rightarrow x \in\left\{0, \frac{\pi}{2}\right\}$ in given range has two solutions

$(II)$

$\left\{x \in\left[\frac{-5 \pi}{18}, \frac{5 \pi}{18}\right]: \sqrt{3} \tan 3 x=1\right\}$

$\sqrt{3} \tan 3 x=1 \Rightarrow \tan 3 x=\frac{1}{\sqrt{3}} \Rightarrow 3 x=n \pi+\frac{\pi}{6}$

$\Rightarrow x=(6 n+1) \frac{\pi}{18} ; n \in Z$

$\Rightarrow x \in\left\{\frac{\pi}{18}, \frac{-5 \pi}{18}\right\}$in given range has two solutions

$\Rightarrow x \in\left\{\frac{\pi}{18}, \frac{-5 \pi}{18}\right\}$ in given range has two solutions

$(III)\left\{x \in\left[-\frac{6 \pi}{5}, \frac{6 \pi}{5}\right]: 2 \cos (2 x)=\sqrt{3}\right\}$

$2 \cos 2 x=\sqrt{3}$

$\Rightarrow \cos 2 x=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$

$\Rightarrow 2 x=2 n \pi \pm \frac{\pi}{6} ; n \in Z$

$\Rightarrow x=n \pi \pm \frac{\pi}{12} ; n \in Z$

$x\left.x \pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12},-\pi \pm \frac{\pi}{12}\right\}$

Six solutions in given range

$(IV) $ $\left\{x \in\left[-\frac{7 \pi}{4}, \frac{7 \pi}{4}\right]: \sin x-\cos x=1\right\}$

$\cos x-\sin x=-1$

$\Rightarrow  \cos \left(x+\frac{\pi}{4}\right)=\frac{-1}{\sqrt{2}}=\cos \frac{3 \pi}{4}$

$\Rightarrow  x+\frac{\pi}{4}=2 n \pi \pm \frac{3 \pi}{4} ; n \in Z$

$\Rightarrow  x=2 n \pi+\frac{\pi}{2} \text { or } x=2 n \pi-\pi ; n \in Z$

$\Rightarrow  x \in\left\{\frac{\pi}{2}, \frac{-3 \pi}{2}, \pi,-\pi\right\}$ four solutions in given range

$=\frac{16 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta) \times 4}{(\cos 2(\alpha-\beta)-\cos 2(\alpha+\beta))^2}$

$=\frac{64 \cos ^4(\alpha-\beta) \sin ^2(\alpha+\beta)}{\left(2 \cos ^2(\alpha-\beta)-1-1+2 \sin ^2(\alpha+\beta)\right)^2}$

$=64 \times \frac{16}{81} \times \frac{1}{9} \frac{1}{\left(2 \times \frac{4}{9}-1-1+\frac{2}{9}\right)^2}$

$=\frac{64 \times 16}{81 \times 9} \cdot \frac{81}{64}=\frac{16}{9}$

${\left[\frac{16}{9}\right]=1 \text { Ans. }}$

$f(\theta)  =\sin ^2 2 \theta-\sin 2 \theta+2$

$f^{\prime}(\theta)  =2(\sin 2 \theta) \cdot(2 \cos 2 \theta)-2 \cos 2 \theta$

$ =2 \cos 2 \theta(2 \sin 2 \theta-1)$

critical points

so, minimum at $\theta=\frac{\pi}{12}, \frac{5 \pi}{12}$

$\lambda_1+\lambda_2=\frac{1}{12}+\frac{5}{12}=\frac{6}{12}=\frac{1}{2}$

$\tan \frac{x}{2}+\tan \frac{z}{2}=\frac{2 y}{x+y+z}$

$\frac{\Delta}{S(S-x)}+\frac{\Delta}{S(S-z)}=\frac{2 y}{2 S}$

$\frac{\Delta}{ S }\left(\frac{2 S -( x + z )}{( S - x )( S - z )}\right)=\frac{ y }{ S }$

$\Rightarrow \frac{\Delta y}{S(S-x)(S-z)}=\frac{y}{S}$

$\Rightarrow \quad \Delta^2=(S-x)^2(S-z)^2$

$\Rightarrow \quad S ( S - y )=( S - x )( S - z )$

$\Rightarrow \quad(x+y+z)(x+z-y)=(y+z-x)(x+y-z)$

$\Rightarrow \quad(x+z)^2-y^2=y^2-(z-x)^2$

$\Rightarrow \quad(x+z)^2+(x-z)^2=2 y^2$

$\Rightarrow \quad x ^2+ z ^2= y ^2 \Rightarrow \angle Y =\frac{\pi}{2}$

$\Rightarrow \quad \angle Y =\angle X +\angle Z$

$\tan \frac{x}{2}=\frac{\Delta}{S(S-x)}$

$\tan \frac{x}{2}=\frac{\frac{1}{2} x z}{\frac{(y+z)^2-x^2}{4}}$

(IMAGE)

$\tan \frac{x}{2}=\frac{2 x z}{y^2+z^2+2 y z-x^2}$

$\tan \frac{x}{2}=\frac{2 x z}{2 z^2+2 y z} \quad \text { (using } y^2=x^2+z^2 \text { ) }$

$\tan \frac{x}{2}=\frac{x}{y+z}$

$X :\{ x : f ( x )=0\}$

$f(x)=0 \Rightarrow \sin (x \cos x)=0 \Rightarrow \cos x=n \Rightarrow \cos x=1,-1,0 \Rightarrow x=\frac{n \pi}{2}$

$x =\left\{\frac{ n \pi }{2}: \pi \in N \right\}-\left\{\frac{\pi}{2}, \pi \cdot \frac{3 \pi}{2}, 2 \pi,\right\}$

$g(x)=\cos (2 \pi \sin x)$

$Z=\{x: g(x)=0\}$

$\cos (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=(2 n+1) \frac{\pi}{2} \Rightarrow \sin x-\frac{(2 n+1)}{4}$

$\sin x=-\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{3}{4}$

$Z=\left\{n \pi \pm \sin ^{-1}\left(\frac{1}{4}\right), m \pi \pm \sin ^{-1}\left(\frac{3}{4}\right), n \in I\right\}$

$Y =\{ x : f ( x )=0\}$

$f(x)=\sin (\pi \cos x) \Rightarrow f^{\prime}(x)=\cos (\pi \cos x) \cdot(-\pi \sin x)=0$

$\sin x=0 \Rightarrow x=m \pi \text {. }$

$\cos (\pi \cos x)=0 \Rightarrow \pi \cos x=(2 n+1) \frac{\pi}{2} \Rightarrow \cos x-\frac{(2 n+1)}{2} \Rightarrow \cos x=-\frac{1}{2} \cdot \frac{1}{2}$

$Y=\left\{2 \pi, n \pi \pm \frac{\pi}{3}\right\}-\left\{\frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}, 2 \pi, \ldots . .\right\}$

$W=\left\{x: g^{\prime}(x)=0\right\}$

$g(x)=\cos (2 \pi \sin x) \Rightarrow g^{\prime}(x)=-\sin (2 \pi \sin x) \cdot(2 \pi \cos x)=0$

$\cos x =0 \Rightarrow x =(2 n +1) \frac{\pi}{2}$

$\sin (2 \pi \sin x)=0 \Rightarrow 2 \pi \sin x=n \pi \Rightarrow \sin x=\frac{\pi}{2}=-1-\frac{1}{2} \cdot 0 \cdot \frac{1}{2} \cdot 1$

$W=\left\{\frac{n \pi}{2}, n \pi \pm \frac{\pi}{6}, \pi \in I\right\}-\left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \pi, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots\right\}$

Now check the options

$\sqrt{3} \cos x+\frac{2 b}{a} \sin x=\frac{c}{a}$

As $\alpha, \beta$ are roots $\Rightarrow \sqrt{3} \cos \alpha+\frac{2 b }{ a } \sin \alpha=\frac{ c }{ a }$    $. . . . . .(1)$

$(1)$ $-(2)$

$\sqrt{3} \cos \beta+\frac{2 b}{a} \sin \beta=\frac{c}{a}$   $. . . . . .(2)$

$\sqrt{3}(\cos \alpha-\cos \beta)+\frac{2 b}{a}(\sin \alpha-\sin \beta)=0$

$\sqrt{3}\left(2 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta-\alpha}{2}\right)+\frac{2 b}{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)=0$

Given $\alpha+\beta=\frac{\pi}{3}$

$\sqrt{3}\left(2 \cdot \frac{1}{2} \sin \frac{\beta-\alpha}{2}\right)+2 \frac{2 b}{a}\left(\frac{\sqrt{3}}{2} \sin \frac{\alpha-\beta}{2}\right)=0$

$\Rightarrow 1=\frac{2 b}{a} \quad \Rightarrow \frac{b}{a}=\frac{1}{2}$

$\frac{\cos \beta+1}{\cos \beta-1}=\frac{3(\cos \alpha+1)}{\cos \alpha-1}$

$\frac{2 \cos ^2 \frac{\beta}{2}}{-2 \sin ^2 \frac{\beta}{2}}=\frac{3 \times 2 \cos ^2 \frac{\alpha}{2}}{-2 \sin ^2 \frac{\alpha}{2}}$

$\tan ^2 \frac{\alpha}{2}=3 \tan ^2 \frac{\beta}{2}$

$\Rightarrow \quad \tan \frac{\alpha}{2}= \pm \sqrt{3} \tan \frac{\beta}{2}$

$\Rightarrow \sqrt{3} \sec x+\csc x=2(\cot x-\tan x)$

$\Rightarrow \frac{(\sqrt{3} \sec x+\csc x)}{2}=\cot x-\tan x$

$\Rightarrow \frac{\sqrt{3}}{2} \sec x +\frac{1}{2} \csc x =\cot x -\tan x$ by dividing both sides by $2$

We know that $\sec x=\frac{1}{\cos x}, \csc x=\frac{1}{\sin x}, \tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$ $\Rightarrow \frac{\sqrt{3} 1}{2 \cos x}+\frac{11}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$

$\Rightarrow \frac{\sqrt{3} \sin x}{2 \sin x \cos x}+\frac{1 \cos x}{2 \sin x \cos x}=\frac{\cos ^2 x}{\sin x \cos x}-\frac{\sin ^2 x}{\cos x \sin x}$

$\Rightarrow \frac{\sqrt{3}}{2} \sin x +\frac{1}{2} \cos x =\cos ^2 x -\sin ^2 x$

$\Rightarrow \sin \frac{\pi}{3} \sin x+\cos \frac{\pi}{3} \cos x=\cos 2 x$ where $\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}$ and $\frac{1}{2}=\cos \frac{\pi}{3}$

$\Rightarrow \cos \left(\frac{\pi}{3}-x\right)=\cos 2 x$

$\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$ since if $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha$

$\Rightarrow 2 x =2 n \pi+ x -\frac{\pi}{3}$ or $2 x =2 n \pi- x +\frac{\pi}{3}$

$\Rightarrow x=2 n \pi-\frac{\pi}{3}$ or $3 x=2 n \pi+\frac{\pi}{3}$ or $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$

For $n =0, x =\frac{-\pi}{3}, \frac{\pi}{9}$

For $n =-1, x =\frac{-7 \pi}{3}$ which does not lie between $(-\pi, \pi)$

and $x=\frac{-2 \pi}{3}+\frac{\pi}{9}=\frac{-5 \pi}{9} \in(-\pi, \pi)$

For $n =2, x =\frac{5 \pi}{3}$ does not lie between $(-\pi, \pi)$

and $x=\frac{2 \pi}{3}+\frac{\pi}{9}=\frac{7 \pi}{9} \in(-\pi, \pi)$

$\therefore x =\frac{-\pi}{3}, \frac{\pi}{9}, \frac{-5 \pi}{9}, \frac{7 \pi}{9}$

Sum of all the solutions $=\frac{-\pi}{3}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=\frac{-3 \pi}{9}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=0$

$\frac{\left(\sin ^2 x\right)^2}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5}$

$\text { Let } \sin ^2 x \text { bet, } t \geq 0$

$t \in[0,1]$

$\frac{ t ^2}{2}+\frac{(1-t)^2}{3}=\frac{1}{5}$

$\frac{ t ^2}{2}+\frac{1-2 t + t ^2}{3}=\frac{1}{5}$

$\frac{3 t ^2+2-4 t +2 t ^2}{6}=\frac{1}{5}$

$25 t ^2-20 t +10=6$

$25 t ^2-20 t +4=0$

$25 t ^2-10 t -10 t +4=0$

$(5 t -2)(5 t -2)=0$

$t =\frac{2}{5}$

$\therefore \sin ^2 x =\frac{2}{5}$

$\cos ^2 x =1-\sin ^2 x =1-\frac{2}{5}=\frac{3}{5}$

$\tan ^2 x =\frac{\sin ^2 x }{\cos ^2 x }=\frac{2}{3}$

$\cos (\alpha - \beta ) = 1$

==> $\alpha - \beta = 0$

==> $\alpha = \beta $$\cos 2\alpha = \frac{1}{e}$

and $ - 2\pi < 2\alpha < 2\pi $

Hence, there will be four solutions.

Combine first two factors, ${f_n}(\theta ) = \frac{{\sin \theta }}{{\cos \theta }}\left[ {\frac{{2{{\cos }^2}\theta }}{{\cos 2\theta }}.\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right]$

Again combine first two factors,

${f_n}(\theta ) = \tan 2\theta \left[ {\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right] = \tan ({2^n}\theta )$

$\therefore {f_2}\left( {\frac{\pi }{{16}}} \right) = \tan \frac{{4\pi }}{{16}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_3}\left( {\frac{\pi }{{32}}} \right) = \tan \frac{{8\pi }}{{32}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_4}\left( {\frac{\pi }{{64}}} \right) = \tan \frac{{16\pi }}{{64}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$

${f_5}\left( {\frac{\pi }{{128}}} \right) = \tan 32\frac{\pi }{{128}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$.

$\cos {15^o} = \cos ({45^o} - {30^o}) = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}=$ irrational 

$\therefore \,\,\,\sin {15^o}\cos {15^o} = \frac{1}{2}(2\sin {15^o}\cos {15^o})$

$ = \frac{1}{2}\sin {30^o} = \frac{1}{2}.\frac{1}{2} = \frac{1}{4} =$ rational  

$\therefore \, \sin {15^o}\cos {75^o} = \sin {15^o}\sin {15^o} = {\sin ^2}{15^o}$ 

$ = {\left( {\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)^2} = \frac{{4 - 2\sqrt 3 }}{8}=$ irrational

==> $\frac{1}{{\cos (\theta - \alpha )}},\frac{1}{{\cos \theta }},\frac{1}{{\cos (\theta + \alpha )}}$ will be in $A.P. $

Hence, $\frac{2}{{\cos \theta }} = \frac{1}{{\cos (\theta - \alpha )}} + \frac{1}{{\cos (\theta + \alpha )}}$ 

$ = \frac{{\cos (\alpha + \theta ) + \cos (\theta - \alpha )}}{{{{\cos }^2}\theta - {{\sin }^2}\alpha }}$ 

==> $\frac{2}{{\cos \theta }} = \frac{{2\cos \theta \cos \alpha }}{{{{\cos }^2}\theta - {{\sin }^2}\alpha }}$

==> ${\cos ^2}\theta - {\sin ^2}\alpha = {\cos ^2}\theta \cos \alpha $

==> ${\cos ^2}\theta \,(1 - \cos \alpha ) = {\sin ^2}\alpha $

==> ${\cos ^2}\theta \left( {2{{\sin }^2}\frac{\alpha }{2}} \right) $

$= 4{\sin ^2}\frac{\alpha }{2}{\cos ^2}\frac{\alpha }{2}$

${\cos ^2}\theta {\sec ^2}\frac{\alpha }{2} = 2$

$\Rightarrow \cos \theta \sec \frac{\alpha }{2} = \pm \sqrt 2 $.

when $\alpha + \beta + \gamma = \pi $. 

$\therefore $ $\sin \alpha + \sin \beta + \sin \gamma > 0$ 

$\therefore $ The minimum value of $\sin \alpha + \sin \beta + \sin \gamma $ is always positive.

$ = \frac{{{{(\cos x - \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}} $

$= \frac{{\cos x - \sin x}}{{\cos x + \sin x}} = \frac{{1 - \tan x}}{{1 + \tan x}}$

$ = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \left( {\frac{\pi }{4}} \right)\sin x}} = \tan \left( {\frac{\pi }{4} - x} \right)$.

$y = 1 + {\sin ^2}\phi + {\sin ^4}\phi + .... = \frac{1}{{(1 - {{\sin }^2}\phi )}} = \frac{1}{{{{\cos }^2}\phi }}$

$z = 1 + {\cos ^2}\phi {\sin ^2}\phi + {\cos ^4}\phi {\sin ^4}\phi + .. = \frac{1}{{(1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

Now $xyz = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}}$

$xy + z = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi }} + \frac{1}{{1 - {{\cos }^2}\phi {{\sin }^2}\phi }}$

$ = \frac{1}{{{{\sin }^2}\phi {{\cos }^2}\phi (1 - {{\cos }^2}\phi {{\sin }^2}\phi )}} = xyz$

which is given in $(b)$

Also $x + y + z = xyz$, which is given in $(c)$.

Hence maximum value will be at $x = \frac{\pi }{{12}}$.

$y = \frac{{1 - 3{{\tan }^2}x}}{{3 - {{\tan }^2}x}} = \frac{{\frac{1}{3} - {{\tan }^2}x}}{{1 - \frac{1}{3}.{{\tan }^2}x}}$ 

Hence, $y$ should never lie between $\frac{1}{3}$ and $3$ whenever defined.